Topology/Sequences

A sequence in a space ${\displaystyle X}$ is a function ${\displaystyle f:\mathbb {N} \to X}$. We write ${\displaystyle a_{n}=f(n)}$ and denote the sequence by ${\displaystyle \langle a_{n}\rangle _{n\in \mathbb {N} }}$ or simply ${\displaystyle (a_{n})}$. Intuitively, a sequence is an infinite list ${\displaystyle a_{1},a_{2},a_{3},\dots }$ of points of ${\displaystyle X}$.

Examples.

1. In ${\displaystyle \mathbb {R} }$, ${\displaystyle a_{n}={\tfrac {1}{n}}}$ is the list ${\displaystyle 1,{\tfrac {1}{2}},{\tfrac {1}{3}},\dots }$.
2. The constant sequence ${\displaystyle a_{n}=1}$ repeats the point ${\displaystyle 1}$ forever.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space and let ${\displaystyle (x_{n})}$ be a sequence in ${\displaystyle X}$. We say ${\displaystyle x_{n}\to x}$ (or that ${\displaystyle (x_{n})}$ converges to ${\displaystyle x}$) if for every neighborhood ${\displaystyle U}$ of ${\displaystyle x}$ there exists ${\displaystyle N\in \mathbb {N} }$ such that ${\displaystyle n\geq N}$ implies ${\displaystyle x_{n}\in U}$. Equivalently, the sequence is eventually in every neighborhood of ${\displaystyle x}$.

Theorem. If ${\displaystyle X}$ is Hausdorff, a sequence of points of ${\displaystyle X}$ has at most one limit.

Proof. Suppose ${\displaystyle x_{n}\to x}$ and ${\displaystyle y\neq x}$. Choose disjoint neighborhoods ${\displaystyle U\ni x}$ and ${\displaystyle V\ni y}$. Since ${\displaystyle x_{n}}$ is eventually in ${\displaystyle U}$, it is not eventually in ${\displaystyle V}$, so it cannot converge to ${\displaystyle y}$. ∎

A subsequence is a sequence ${\displaystyle x_{n_{k}}}$ with strictly increasing indices ${\displaystyle n_{1}<n_{2}<\dots }$. If ${\displaystyle x_{n}\to x}$ then every subsequence also converges to ${\displaystyle x}$.

A point ${\displaystyle x}$ is a cluster point (or limit point) of ${\displaystyle (x_{n})}$ if every neighborhood ${\displaystyle U\ni x}$ contains infinitely many terms of the sequence. In first-countable spaces (see below), ${\displaystyle x}$ is a cluster point of ${\displaystyle (x_{n})}$ iff some subsequence converges to ${\displaystyle x}$.

A space ${\displaystyle X}$ is first countable if each ${\displaystyle x\in X}$ has a countable neighborhood base ${\displaystyle \{U_{k}\}_{k\in \mathbb {N} }}$ with ${\displaystyle U_{k+1}\subseteq U_{k}}$. Metric spaces are first countable.

Theorem (Sequential characterization of closure in first-countable spaces).

If ${\displaystyle X}$ is first countable and ${\displaystyle A\subseteq X}$, then ${\displaystyle x\in {\overline {A}}}$ iff there exists a sequence ${\displaystyle (a_{n})}$ in ${\displaystyle A}$ with ${\displaystyle a_{n}\to x}$.

Proof. If ${\displaystyle x\in {\overline {A}}}$, pick ${\displaystyle a_{n}\in A\cap U_{n}}$ from a nested countable neighborhood base ${\displaystyle (U_{n})}$ at ${\displaystyle x}$. Conversely, if ${\displaystyle a_{n}\to x}$, then every neighborhood of ${\displaystyle x}$ contains eventually all ${\displaystyle a_{n}}$, hence meets ${\displaystyle A}$.

In this section we use the topology on ${\displaystyle \mathbb {R} }$ having as base all sets of the form

${\displaystyle O\setminus C,}$

where ${\displaystyle O}$ is open in the usual Euclidean topology on ${\displaystyle \mathbb {R} }$ and ${\displaystyle C}$ is countable. This is sometimes called the co-countable extension topology (it refines the usual topology by also including sets obtained by removing countably many points from usual opens). In particular, every usual open set ${\displaystyle O}$ remains open because ${\displaystyle O=O\setminus \varnothing }$ is a basic open here. Hence this topology is finer than the usual one.

• A sequence ${\displaystyle (x_{n})}$ converges to ${\displaystyle x}$ iff it is eventually constant at ${\displaystyle x}$.

Proof. Suppose ${\displaystyle x_{n}\to x}$ in this topology. Let

${\displaystyle C=\{x_{n}:\;x_{n}\neq x\}}$

the (at most) countable set of values attained by the sequence that differ from ${\displaystyle x}$. Then ${\displaystyle \mathbb {R} \setminus C}$ is a basic open neighborhood of ${\displaystyle x}$. By convergence, there is ${\displaystyle N}$ such that for all ${\displaystyle n\geq N}$, ${\displaystyle x_{n}\in \mathbb {R} \setminus C}$. But for any term of the sequence, either ${\displaystyle x_{n}=x}$ or ${\displaystyle x_{n}\in C}$ by definition of ${\displaystyle C}$. Thus ${\displaystyle x_{n}\in \mathbb {R} \setminus C}$ forces ${\displaystyle x_{n}=x}$. Hence the sequence is eventually constant at ${\displaystyle x}$.

Conversely, if ${\displaystyle x_{n}=x}$ for all ${\displaystyle n\geq N}$, then for any neighborhood ${\displaystyle V\ni x}$ we have ${\displaystyle x_{n}\in V}$ for all ${\displaystyle n\geq N}$, so ${\displaystyle x_{n}\to x}$.

• ${\displaystyle {\overline {(0,1)}}=[0,1]}$.

Proof. Because this topology is finer than the usual topology, every set that is closed in the usual topology is also closed here: if ${\displaystyle C=X\setminus O}$ with ${\displaystyle O}$ usual-open, then ${\displaystyle O}$ is still open in the refined topology, hence ${\displaystyle C}$ remains closed. In particular, ${\displaystyle [0,1]}$ is closed, so the closure of ${\displaystyle (0,1)}$ is contained in ${\displaystyle [0,1]}$.

To see that ${\displaystyle 0}$ and ${\displaystyle 1}$ lie in the closure, let ${\displaystyle x\in \{0,1\}}$ and let ${\displaystyle U}$ be any neighborhood of ${\displaystyle x}$ in the refined topology. Choose a basic neighborhood ${\displaystyle V=O\setminus C}$ with ${\displaystyle x\in V\subseteq U}$, where ${\displaystyle O}$ is usual-open and ${\displaystyle C}$ is countable. Since every usual-open ${\displaystyle O}$ containing ${\displaystyle x}$ meets ${\displaystyle (0,1)}$ in an (uncountable) interval, ${\displaystyle (O\cap (0,1))\setminus C\neq \varnothing }$. Thus ${\displaystyle V\cap (0,1)\neq \varnothing }$, and therefore ${\displaystyle U\cap (0,1)\neq \varnothing }$. Hence each of ${\displaystyle 0}$ and ${\displaystyle 1}$ is a closure point of ${\displaystyle (0,1)}$, giving ${\displaystyle {\overline {(0,1)}}=[0,1]}$.

A directed set is a set ${\displaystyle D}$ with a reflexive, transitive relation ${\displaystyle \geq }$ such that for any ${\displaystyle \alpha ,\beta \in D}$ there exists ${\displaystyle \gamma \in D}$ with ${\displaystyle \gamma \geq \alpha }$ and ${\displaystyle \gamma \geq \beta }$ (an upper bound). Examples:

• ${\displaystyle \mathbb {N} }$ with the usual order;
• ${\displaystyle (0,\infty )}$ with ${\displaystyle x\geq y\iff x\geq y}$;
• the neighborhood system ${\displaystyle {\mathcal {N}}_{x}}$ at a point ${\displaystyle x}$, directed by reverse inclusion: ${\displaystyle V\geq W\iff V\subseteq W}$;
• finite-subset filter base: all finite ${\displaystyle \Phi \subseteq X}$ directed by reverse inclusion ${\displaystyle A\geq B\iff A\subseteq B}$.

Products of directed sets are directed by the product order.

A net in ${\displaystyle X}$ is a function ${\displaystyle x:D\to X}$ from a directed set ${\displaystyle D}$. We write ${\displaystyle (x_{\alpha })_{\alpha \in D}}$. A net converges to ${\displaystyle x\in X}$ if for every neighborhood ${\displaystyle V\ni x}$ there exists ${\displaystyle \alpha _{0}}$ such that ${\displaystyle \alpha \geq \alpha _{0}}$ implies ${\displaystyle x_{\alpha }\in V}$.

Theorem. Let ${\displaystyle X}$ be a topological space and ${\displaystyle A\subseteq X}$. A point ${\displaystyle x\in X}$ satisfies ${\displaystyle x\in {\overline {A}}}$ iff there exists a net ${\displaystyle (x_{\alpha })}$ with each ${\displaystyle x_{\alpha }\in A}$ and ${\displaystyle x_{\alpha }\to x}$.

Proof. (${\displaystyle \Rightarrow }$) Assume ${\displaystyle x\in {\overline {A}}}$. Let ${\displaystyle {\mathcal {N}}_{x}}$ be the set of neighborhoods of ${\displaystyle x}$, directed by reverse inclusion: ${\displaystyle V\geq W\iff V\subseteq W}$. For each ${\displaystyle V\in {\mathcal {N}}_{x}}$ we have ${\displaystyle V\cap A\neq \varnothing }$; choose ${\displaystyle a_{V}\in V\cap A}$. Then ${\displaystyle (a_{V})_{V\in {\mathcal {N}}_{x}}}$ is a net in ${\displaystyle A}$. It converges to ${\displaystyle x}$: given any neighborhood ${\displaystyle W\ni x}$, for all ${\displaystyle V\geq W}$ we have ${\displaystyle V\subseteq W}$, hence ${\displaystyle a_{V}\in V\subseteq W}$.

(${\displaystyle \Leftarrow }$) Conversely, suppose ${\displaystyle (x_{\alpha })}$ is a net in ${\displaystyle A}$ with ${\displaystyle x_{\alpha }\to x}$. If ${\displaystyle U}$ is any neighborhood of ${\displaystyle x}$, then ${\displaystyle x_{\alpha }\in U}$ eventually, so ${\displaystyle U\cap A\neq \varnothing }$. Hence every neighborhood of ${\displaystyle x}$ meets ${\displaystyle A}$, i.e. ${\displaystyle x\in {\overline {A}}}$. ∎

Remark (choice-free variant). It can be noted that this result doesn't quite depend on the axiom of choice because we can avoid the selection by indexing by ${\displaystyle D=\{(V,a):V\in {\mathcal {N}}_{x},\ a\in V\cap A\}}$ with ${\displaystyle (V,a)\leq (W,b)}$ iff ${\displaystyle W\subseteq V}$; define the net ${\displaystyle y_{(V,a)}=a}$. Then ${\displaystyle y_{(V,a)}\to x}$ for the same reason as above. All the following theorems in the net section can be modified in a similar manner to avoid using choice, which can be an exercise.

Theorem (Hausdorff ⇔ unique net limits). A topological space ${\displaystyle X}$ is Hausdorff if and only if every net in ${\displaystyle X}$ converges to at most one point.

Proof. (${\displaystyle \Rightarrow }$) If ${\displaystyle X}$ is Hausdorff and a net ${\displaystyle (x_{\alpha })}$ converges to both ${\displaystyle x}$ and ${\displaystyle y}$ with ${\displaystyle x\neq y}$, choose disjoint neighborhoods ${\displaystyle U\ni x}$ and ${\displaystyle V\ni y}$. Being eventually in ${\displaystyle U}$ and eventually in ${\displaystyle V}$ forces the net to be eventually in ${\displaystyle U\cap V=\varnothing }$, impossible. Hence limits are unique.

(${\displaystyle \Leftarrow }$) Suppose every net in ${\displaystyle X}$ has at most one limit, yet ${\displaystyle X}$ is not Hausdorff. Then there exist distinct points ${\displaystyle x\neq y}$ such that every neighborhood ${\displaystyle U}$ of ${\displaystyle x}$ meets every neighborhood ${\displaystyle V}$ of ${\displaystyle y}$. Consider the directed sets ${\displaystyle {\mathcal {N}}_{x}}$ and ${\displaystyle {\mathcal {N}}_{y}}$ of neighborhoods of ${\displaystyle x}$ and ${\displaystyle y}$, ordered by reverse inclusion: ${\displaystyle U\geq U'\iff U\subseteq U'}$. Their product ${\displaystyle {\mathcal {N}}_{x}\times {\mathcal {N}}_{y}}$ is directed by ${\displaystyle (U,V)\geq (U',V')\iff U\subseteq U',\ V\subseteq V'}$.

For each ${\displaystyle (U,V)}$, choose a point ${\displaystyle x_{U,V}\in U\cap V}$ (nonempty by assumption). Then the net ${\displaystyle {\bigl (}x_{U,V}{\bigr )}_{(U,V)\in {\mathcal {N}}_{x}\times {\mathcal {N}}_{y}}}$ converges to ${\displaystyle x}$ (given a neighborhood ${\displaystyle U_{0}\ni x}$, take ${\displaystyle (U_{0},V)}$ for any V) and also to ${\displaystyle y}$ (symmetrically). This contradicts uniqueness of limits for nets. Therefore ${\displaystyle X}$ must be Hausdorff. ∎

We say ${\displaystyle x}$ is a cluster point (or limit point) of a net ${\displaystyle (x_{\alpha })_{\alpha \in A}}$ if for every neighborhood ${\displaystyle V\ni x}$ and every ${\displaystyle \alpha _{0}\in A}$ there exists ${\displaystyle \beta \geq \alpha _{0}}$ with ${\displaystyle x_{\beta }\in V}$.

A net ${\displaystyle (y_{\lambda })_{\lambda \in \Lambda }}$ is a subnet of ${\displaystyle (x_{\alpha })_{\alpha \in A}}$ if there exists a function ${\displaystyle \varphi :\Lambda \to A}$ such that:

1. ${\displaystyle y_{\lambda }=x_{\varphi (\lambda )}}$ for all ${\displaystyle \lambda \in \Lambda }$; and
2. for every ${\displaystyle \alpha _{0}\in A}$ there exists ${\displaystyle \lambda _{0}\in \Lambda }$ with ${\displaystyle \lambda \geq \lambda _{0}\Rightarrow \varphi (\lambda )\geq \alpha _{0}}$. (This is often described as the image of ${\displaystyle \varphi }$ being cofinal in ${\displaystyle A}$.)

Theorem (Cluster point ⇔ limit of a subnet). Let ${\displaystyle (x_{\alpha })_{\alpha \in A}}$ be a net in a topological space ${\displaystyle X}$ and let ${\displaystyle x\in X}$. Then ${\displaystyle x}$ is a cluster point of ${\displaystyle (x_{\alpha })}$ if and only if there exists a subnet of ${\displaystyle (x_{\alpha })}$ that converges to ${\displaystyle x}$.

Proof. (${\displaystyle \Rightarrow }$) Assume ${\displaystyle x}$ is a cluster point. Consider the directed set ${\displaystyle A\times {\mathcal {N}}_{x}}$, ordered by the product of the given order on ${\displaystyle A}$ and reverse inclusion on ${\displaystyle {\mathcal {N}}_{x}}$:

${\displaystyle (\alpha ,V)\leq (\alpha ',V')\quad \Longleftrightarrow \quad \alpha \leq \alpha '\ {\text{ and }}\ V'\subseteq V.}$

For each ${\displaystyle (\alpha ,V)}$, by the cluster property choose ${\displaystyle \beta (\alpha ,V)\geq \alpha }$ with ${\displaystyle x_{\beta (\alpha ,V)}\in V}$. Define

${\displaystyle y_{(\alpha ,V)}:=x_{\beta (\alpha ,V)}\quad {\text{and}}\quad \varphi (\alpha ,V):=\beta (\alpha ,V).}$

Then ${\displaystyle (y_{(\alpha ,V)})}$ is a subnet of ${\displaystyle (x_{\alpha })}$: condition (1) is by definition; for (2), given ${\displaystyle \alpha _{0}}$ take ${\displaystyle \lambda _{0}=(\alpha _{0},V)}$ (any ${\displaystyle V\in {\mathcal {N}}_{x}}$); if ${\displaystyle (\alpha ,V)\geq (\alpha _{0},V)}$, then ${\displaystyle \alpha \geq \alpha _{0}}$, hence ${\displaystyle \varphi (\alpha ,V)=\beta (\alpha ,V)\geq \alpha \geq \alpha _{0}}$. Finally, ${\displaystyle y_{(\alpha ,V)}\to x}$: for any neighborhood ${\displaystyle W\ni x}$, if ${\displaystyle (\alpha ,V)\geq (\alpha _{0},W)}$ then ${\displaystyle V\subseteq W}$ and ${\displaystyle y_{(\alpha ,V)}\in V\subseteq W}$.

(${\displaystyle \Leftarrow }$) Suppose a subnet ${\displaystyle y_{\lambda }=x_{\varphi (\lambda )}}$ converges to ${\displaystyle x}$. Let ${\displaystyle V}$ be a neighborhood of ${\displaystyle x}$ and ${\displaystyle \alpha _{0}\in A}$. By convergence there exists ${\displaystyle \lambda _{0}}$ such that ${\displaystyle \lambda \geq \lambda _{0}\Rightarrow y_{\lambda }\in V}$. By cofinality, there exists ${\displaystyle \lambda _{1}}$ such that ${\displaystyle \lambda \geq \lambda _{1}\Rightarrow \varphi (\lambda )\geq \alpha _{0}}$. There exists ${\displaystyle \lambda \geq \lambda _{0},\lambda _{1}}$, and for this, ${\displaystyle x_{\varphi (\lambda )}=y_{\lambda }\in V}$ and ${\displaystyle \varphi (\lambda )\geq \alpha _{0}}$, proving the cluster property. ∎

Corollary (Convergence is inherited by subnets and detected by them). A net ${\displaystyle (x_{\alpha })}$ converges to ${\displaystyle x}$ if and only if every subnet converges to ${\displaystyle x}$.

Proof. If ${\displaystyle x_{\alpha }\to x}$, then any subnet is eventually in every neighborhood of ${\displaystyle x}$ because of the cofinality requirement. Conversely, any net is a subnet of itself, so if all subnets converge to ${\displaystyle x}$, then the net itself converges to ${\displaystyle x}$. ∎

A filter on a set ${\displaystyle X}$ is a nonempty family ${\displaystyle {\mathcal {F}}\subseteq {\mathcal {P}}(X)}$ such that:

1. ${\displaystyle \varnothing \notin {\mathcal {F}}}$ and ${\displaystyle X\in {\mathcal {F}}}$;
2. if ${\displaystyle A,B\in {\mathcal {F}}}$, then ${\displaystyle A\cap B\in {\mathcal {F}}}$;
3. if ${\displaystyle A\in {\mathcal {F}}}$ and ${\displaystyle A\subseteq B\subseteq X}$, then ${\displaystyle B\in {\mathcal {F}}}$.

A filter ${\displaystyle {\mathcal {G}}}$ is a subfilter (i.e., is finer) than ${\displaystyle {\mathcal {F}}}$ if ${\displaystyle {\mathcal {F}}\subseteq {\mathcal {G}}}$.

The neighborhood filter at ${\displaystyle x\in X}$ is ${\displaystyle {\mathcal {N}}_{x}=\{U\subseteq X:U{\text{ is a neighborhood of }}x\}}$.

A filter is fixed if ${\displaystyle \bigcap {\mathcal {F}}\neq \varnothing }$ and free otherwise.

A nonempty family ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {P}}(X)}$ is a filter base iff:

1. ${\displaystyle \varnothing \notin {\mathcal {B}}}$; and
2. (directed by reverse inclusion) for all ${\displaystyle B_{1},B_{2}\in {\mathcal {B}}}$ there exists ${\displaystyle C\in {\mathcal {B}}}$ with ${\displaystyle C\subseteq B_{1}\cap B_{2}}$.

Given a filter base ${\displaystyle {\mathcal {B}}}$, its generated filter is ${\displaystyle {\mathcal {F}}_{\mathcal {B}}=\{A\subseteq X:\ \exists B\in {\mathcal {B}}{\text{ with }}B\subseteq A\}.}$

Lemma. ${\displaystyle {\mathcal {F}}_{\mathcal {B}}}$ is a filter, and it is the smallest filter containing ${\displaystyle {\mathcal {B}}}$ (i.e., if ${\displaystyle {\mathcal {F}}}$ is a filter with ${\displaystyle {\mathcal {B}}\subseteq {\mathcal {F}}}$, then ${\displaystyle {\mathcal {F}}_{\mathcal {B}}\subseteq {\mathcal {F}}}$).

Proof.

• Nonemptiness and ${\displaystyle \varnothing \notin {\mathcal {F}}_{\mathcal {B}}}$: since ${\displaystyle {\mathcal {B}}\neq \varnothing }$ and ${\displaystyle \varnothing \notin {\mathcal {B}}}$, every ${\displaystyle B\in {\mathcal {B}}}$ witnesses ${\displaystyle X\in {\mathcal {F}}_{\mathcal {B}}}$, while ${\displaystyle \varnothing \notin {\mathcal {F}}_{\mathcal {B}}}$ because no ${\displaystyle B\subseteq \varnothing }$.
• Upward closure: if ${\displaystyle A\in {\mathcal {F}}_{\mathcal {B}}}$ via ${\displaystyle B\subseteq A}$ and ${\displaystyle A\subseteq A'}$, then ${\displaystyle B\subseteq A'}$, so ${\displaystyle A'\in {\mathcal {F}}_{\mathcal {B}}}$.
• Finite intersections: if ${\displaystyle A_{1},A_{2}\in {\mathcal {F}}_{\mathcal {B}}}$ via ${\displaystyle B_{1}\subseteq A_{1}}$ and ${\displaystyle B_{2}\subseteq A_{2}}$, choose ${\displaystyle C\in {\mathcal {B}}}$ with ${\displaystyle C\subseteq B_{1}\cap B_{2}}$. Then ${\displaystyle C\subseteq A_{1}\cap A_{2}}$, hence ${\displaystyle A_{1}\cap A_{2}\in {\mathcal {F}}_{\mathcal {B}}}$.

Minimality is immediate from the definition. ∎

An ultrafilter is a maximal filter under inclusion.

Ultrafilter Lemma. Every filter on ${\displaystyle X}$ is contained in an ultrafilter.

Proof. Let ${\displaystyle {\mathcal {F}}}$ be a filter. Partially order ${\displaystyle {\mathcal {C}}=\{{\mathcal {G}}:{\mathcal {G}}{\text{ is a filter and }}{\mathcal {F}}\subseteq {\mathcal {G}}\}}$ by inclusion. Any chain ${\displaystyle {\mathcal {A}}\subseteq {\mathcal {C}}}$ has an upper bound ${\displaystyle {\mathcal {H}}=\{A\subseteq X:\exists {\mathcal {G}}\in {\mathcal {A}}{\text{ with }}A\in {\mathcal {G}}\}}$, which is a filter containing all members of the chain. By Zorn’s lemma, ${\displaystyle {\mathcal {C}}}$ has a maximal element, i.e., an ultrafilter containing ${\displaystyle {\mathcal {F}}}$. ∎

Corollary. If ${\displaystyle X}$ is infinite, ${\displaystyle \{A\subseteq X:A^{c}{\text{ is finite}}\}}$ extends to a (free) ultrafilter.

For ${\displaystyle x\in X}$, the principal (or point) ultrafilter at ${\displaystyle x}$ is ${\displaystyle {\mathcal {U}}_{x}=\{A\subseteq X:x\in A\}}$.

Proposition. Every fixed ultrafilter ${\displaystyle {\mathcal {U}}}$ on ${\displaystyle X}$ equals ${\displaystyle {\mathcal {U}}_{x}}$ for a unique ${\displaystyle x\in X}$.

Proof. Since ${\displaystyle {\mathcal {U}}}$ is fixed, pick ${\displaystyle x\in \bigcap {\mathcal {U}}}$. Then ${\displaystyle {\mathcal {U}}\subseteq {\mathcal {U}}_{x}}$, and maximality yields ${\displaystyle {\mathcal {U}}={\mathcal {U}}_{x}}$. If ${\displaystyle {\mathcal {U}}_{x}={\mathcal {U}}_{y}}$, then ${\displaystyle \{x\}\in {\mathcal {U}}_{y}}$ so ${\displaystyle y\in \{x\}}$ and ${\displaystyle x=y}$. ∎

Lemma (finite-union property). If ${\displaystyle {\mathcal {U}}}$ is an ultrafilter and ${\displaystyle A\cup B\in {\mathcal {U}}}$, then ${\displaystyle A\in {\mathcal {U}}}$ or ${\displaystyle B\in {\mathcal {U}}}$. Consequently, if ${\displaystyle A_{1}\cup \cdots \cup A_{n}\in {\mathcal {U}}}$, then ${\displaystyle A_{i}\in {\mathcal {U}}}$ for some ${\displaystyle i}$.

Proof. Define ${\displaystyle {\mathcal {F}}=\{C\subseteq X:A\cup C\in {\mathcal {U}}\}}$. Then ${\displaystyle {\mathcal {F}}}$ is a filter, ${\displaystyle {\mathcal {U}}\subseteq {\mathcal {F}}}$, and ${\displaystyle B\in {\mathcal {F}}}$. By maximality, ${\displaystyle {\mathcal {F}}={\mathcal {U}}}$, hence ${\displaystyle B\in {\mathcal {U}}}$. The ${\displaystyle n}$-ary case follows by induction. ∎

Lemma (hitting every member forces membership). If ${\displaystyle {\mathcal {U}}}$ is an ultrafilter and ${\displaystyle A\cap B\neq \varnothing }$ for all ${\displaystyle B\in {\mathcal {U}}}$, then ${\displaystyle A\in {\mathcal {U}}}$.

Proof. The family ${\displaystyle {\mathcal {B}}=\{A\cap B:B\in {\mathcal {U}}\}}$ is a filter base (nonempty and directed). Let ${\displaystyle {\mathcal {F}}_{\mathcal {B}}}$ be its generated filter. Then ${\displaystyle {\mathcal {U}}\subseteq {\mathcal {F}}_{\mathcal {B}}}$ and ${\displaystyle A\in {\mathcal {F}}_{\mathcal {B}}}$. Maximality forces ${\displaystyle {\mathcal {F}}_{\mathcal {B}}={\mathcal {U}}}$, so ${\displaystyle A\in {\mathcal {U}}}$. ∎

Corollary (dichotomy). For any ${\displaystyle A\subseteq X}$ and ultrafilter ${\displaystyle {\mathcal {U}}}$, exactly one of ${\displaystyle A}$ or ${\displaystyle A^{c}}$ lies in ${\displaystyle {\mathcal {U}}}$.

Proof. Apply the finite-union property to ${\displaystyle A\cup A^{c}=X\in {\mathcal {U}}}$ and note that ${\displaystyle \varnothing \notin {\mathcal {U}}}$. ∎

Proposition. If ${\displaystyle {\mathcal {U}}}$ is a free ultrafilter on ${\displaystyle X}$, then ${\displaystyle {\mathcal {U}}}$ contains no finite subset of ${\displaystyle X}$.

Proof. Suppose ${\displaystyle F=\{x_{1},\dots ,x_{n}\}\in {\mathcal {U}}}$. By the finite-union property, ${\displaystyle \{x_{i}\}\in {\mathcal {U}}}$ for some ${\displaystyle i}$, which makes ${\displaystyle {\mathcal {U}}}$ fixed (indeed ${\displaystyle {\mathcal {U}}={\mathcal {U}}_{x_{i}}}$), a contradiction. ∎

Consequences. Only infinite sets admit free ultrafilters; every finite-set ultrafilter is principal.

For a filter ${\displaystyle {\mathcal {F}}}$ on a topological space ${\displaystyle X}$:

• Convergence. ${\displaystyle {\mathcal {F}}}$ converges to ${\displaystyle x}$ (write ${\displaystyle {\mathcal {F}}\to x}$) iff ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {F}}}$, where ${\displaystyle {\mathcal {N}}_{x}}$ is the neighborhood filter at ${\displaystyle x}$.
• Adherence (limit points). The adherence set (set of limit/cluster points) of ${\displaystyle {\mathcal {F}}}$ is

${\displaystyle \operatorname {Adh} {\mathcal {F}}=\bigcap _{A\in {\mathcal {F}}}{\overline {A}}.}$

Equivalently, ${\displaystyle x\in \operatorname {Adh} {\mathcal {F}}}$ iff every neighborhood of ${\displaystyle x}$ meets every ${\displaystyle A\in {\mathcal {F}}}$.

Note. Some texts reserve “limit” for convergence (${\displaystyle {\mathcal {F}}\to x}$) and use “adherent/cluster point” for ${\displaystyle \operatorname {Adh} {\mathcal {F}}}$. We adopt this distinction here.

If ${\displaystyle {\mathcal {F}}\to x}$, then ${\displaystyle x\in \operatorname {Adh} {\mathcal {F}}}$ (trivial since ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {F}}}$ implies every ${\displaystyle A\in {\mathcal {F}}}$ meets every ${\displaystyle V\in {\mathcal {N}}_{x}}$).

Theorem. A space ${\displaystyle X}$ is Hausdorff iff every filter on ${\displaystyle X}$ has at most one limit (i.e., if ${\displaystyle {\mathcal {F}}\to x}$ and ${\displaystyle {\mathcal {F}}\to y}$, then ${\displaystyle x=y}$).

Proof. (${\displaystyle \Rightarrow }$) Suppose ${\displaystyle X}$ is Hausdorff and ${\displaystyle {\mathcal {F}}\to x}$ and ${\displaystyle {\mathcal {F}}\to y}$. Pick disjoint neighborhoods ${\displaystyle U\in {\mathcal {N}}_{x}}$, ${\displaystyle V\in {\mathcal {N}}_{y}}$. Then ${\displaystyle U,V\in {\mathcal {F}}}$, so ${\displaystyle U\cap V\in {\mathcal {F}}}$. But ${\displaystyle U\cap V=\varnothing }$, contradicting ${\displaystyle \varnothing \notin {\mathcal {F}}}$. Hence ${\displaystyle x=y}$.

(${\displaystyle \Leftarrow }$) Assume every filter has a unique limit; suppose toward a contradiction that ${\displaystyle X}$ is not Hausdorff. Then there exist distinct ${\displaystyle x\neq y}$ such that no neighborhoods of ${\displaystyle x}$ and ${\displaystyle y}$ are disjoint (equivalently, ${\displaystyle U\cap V\neq \varnothing }$ for all ${\displaystyle U\in {\mathcal {N}}_{x}}$, ${\displaystyle V\in {\mathcal {N}}_{y}}$). Consider the base

${\displaystyle {\mathcal {B}}=\{U\cap V:\ U\in {\mathcal {N}}_{x},\ V\in {\mathcal {N}}_{y}\}.}$

This is a filter base: ${\displaystyle \varnothing \notin {\mathcal {B}}}$ by the nonseparation assumption, and it is directed since ${\displaystyle (U_{1}\cap V_{1})\cap (U_{2}\cap V_{2})=(U_{1}\cap U_{2})\cap (V_{1}\cap V_{2})}$ with both ${\displaystyle U_{1}\cap U_{2}\in {\mathcal {N}}_{x}}$ and ${\displaystyle V_{1}\cap V_{2}\in {\mathcal {N}}_{y}}$. Let ${\displaystyle {\mathcal {G}}={\mathcal {F}}_{\mathcal {B}}}$ be the generated filter. Then

• For any ${\displaystyle U_{0}\in {\mathcal {N}}_{x}}$, since ${\displaystyle X\in {\mathcal {N}}_{y}}$ we have ${\displaystyle U_{0}=U_{0}\cap X\in {\mathcal {B}}\subseteq {\mathcal {G}}}$. Thus ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {G}}}$ and ${\displaystyle {\mathcal {G}}\to x}$.
• Symmetrically, ${\displaystyle {\mathcal {N}}_{y}\subseteq {\mathcal {G}}}$ (use ${\displaystyle X\in {\mathcal {N}}_{x}}$), so ${\displaystyle {\mathcal {G}}\to y}$.

Hence ${\displaystyle {\mathcal {G}}}$ has two distinct limits, contradicting uniqueness. Therefore ${\displaystyle X}$ must be Hausdorff. ∎

Theorem. A point ${\displaystyle x\in X}$ belongs to ${\displaystyle \operatorname {Adh} {\mathcal {F}}}$ iff there exists a subfilter ${\displaystyle {\mathcal {G}}\supseteq {\mathcal {F}}}$ with ${\displaystyle {\mathcal {G}}\to x}$.

Proof. (${\displaystyle \Rightarrow }$) Assume ${\displaystyle x\in \bigcap _{A\in {\mathcal {F}}}{\overline {A}}}$. For each ${\displaystyle V\in {\mathcal {N}}_{x}}$ and ${\displaystyle A\in {\mathcal {F}}}$, ${\displaystyle V\cap A\neq \varnothing }$. The family ${\displaystyle {\mathcal {B}}=\{V\cap A:\ V\in {\mathcal {N}}_{x},\ A\in {\mathcal {F}}\}}$ is a filter base (directed by reverse inclusion). Let ${\displaystyle {\mathcal {G}}}$ be the filter it generates. Then ${\displaystyle {\mathcal {F}}\subseteq {\mathcal {G}}}$ and ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {G}}}$, hence ${\displaystyle {\mathcal {G}}\to x}$.

(${\displaystyle \Leftarrow }$) If ${\displaystyle {\mathcal {G}}\supseteq {\mathcal {F}}}$ and ${\displaystyle {\mathcal {G}}\to x}$, then for each ${\displaystyle A\in {\mathcal {F}}}$ and each ${\displaystyle V\in {\mathcal {N}}_{x}}$ we have ${\displaystyle V\cap A\in {\mathcal {G}}}$, so ${\displaystyle V\cap A\neq \varnothing }$. Thus ${\displaystyle x\in {\overline {A}}}$ for all ${\displaystyle A\in {\mathcal {F}}}$, i.e. ${\displaystyle x\in \operatorname {Adh} {\mathcal {F}}}$. ∎

Theorem. A filter ${\displaystyle {\mathcal {F}}}$ converges to ${\displaystyle x}$ iff every subfilter ${\displaystyle {\mathcal {G}}\supseteq {\mathcal {F}}}$ converges to ${\displaystyle x}$.

Proof. (${\displaystyle \Rightarrow }$) If ${\displaystyle {\mathcal {F}}\to x}$ and ${\displaystyle {\mathcal {G}}\supseteq {\mathcal {F}}}$, then ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {F}}\subseteq {\mathcal {G}}}$, hence ${\displaystyle {\mathcal {G}}\to x}$.

(${\displaystyle \Leftarrow }$) Every filter is a subfilter of itself, so if every subfilter converged to ${\displaystyle x}$, then the filter converges to ${\displaystyle x}$

Let ${\displaystyle (x_{\alpha })_{\alpha \in D}}$ be a net in a topological space ${\displaystyle X}$, where ${\displaystyle D}$ is a directed set.

For each ${\displaystyle \alpha \in D}$, define the tail set

${\displaystyle F_{\alpha }=\{x_{\beta }:\ \beta \geq \alpha \}\ \subseteq X.}$

The family ${\displaystyle {\mathcal {B}}=\{F_{\alpha }:\alpha \in D\}}$ is a filter base on ${\displaystyle X}$:

• ${\displaystyle \varnothing \notin {\mathcal {B}}}$ since ${\displaystyle x_{\alpha }\in F_{\alpha }}$;
• for ${\displaystyle \alpha ,\alpha '}$ choose ${\displaystyle \gamma \geq \alpha ,\alpha '}$; then ${\displaystyle F_{\gamma }\subseteq F_{\alpha }\cap F_{\alpha '}}$.

The section filter (or tail filter) of ${\displaystyle (x_{\alpha })}$) is the filter generated by ${\displaystyle {\mathcal {B}}}$, denoted ${\displaystyle {\mathcal {F}}_{x}={\mathcal {F}}_{\{F_{\alpha }\}}}$.

The adherence set of a filter ${\displaystyle {\mathcal {F}}}$ is

${\displaystyle \operatorname {Adh} {\mathcal {F}}=\bigcap _{A\in {\mathcal {F}}}{\overline {A}}.}$

Given a filter ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle X}$, define the directed set

${\displaystyle D_{\mathcal {F}}=\{(A,a):\ A\in {\mathcal {F}},\ a\in A\}\quad {\text{with}}\quad (A,a)\leq (B,b)\iff B\subseteq A}$

(reverse inclusion on the set coordinate). The net generated by ${\displaystyle {\mathcal {F}}}$ is

${\displaystyle x_{(A,a)}=a,\qquad (A,a)\in D_{\mathcal {F}}.}$

Lemma 1 (subnet ⇒ subfilter). If ${\displaystyle (y_{\lambda })_{\lambda \in \Lambda }}$ is a subnet of ${\displaystyle (x_{\alpha })_{\alpha \in D}}$ (i.e., ${\displaystyle y_{\lambda }=x_{\varphi (\lambda )}}$ with ${\displaystyle \varphi :\Lambda \to D}$ cofinal), then [ \mathcal F_y\ \subseteq\ \mathcal F_x . ]

Proof. For some ${\displaystyle \lambda _{0}}$ and every ${\displaystyle \lambda \geq \lambda _{0}}$, cofinality gives ${\displaystyle \varphi (\lambda )\geq \varphi (\lambda _{0})}$. Thus the tail ${\displaystyle F_{\lambda _{0}}^{(y)}=\{y_{\lambda }:\lambda \geq \lambda _{0}\}}$ is contained in the tail ${\displaystyle F_{\varphi (\lambda _{0})}^{(x)}}$. Hence every base element of ${\displaystyle {\mathcal {F}}_{y}}$ is contained in a base element of ${\displaystyle {\mathcal {F}}_{x}}$, so the generated filter ${\displaystyle {\mathcal {F}}_{y}}$ is a subfilter of ${\displaystyle {\mathcal {F}}_{x}}$. ∎

Lemma 2 (net convergence ⇔ section filter convergence). A net ${\displaystyle (x_{\alpha })}$ converges to ${\displaystyle x}$ iff its section filter ${\displaystyle {\mathcal {F}}_{x}}$ converges to ${\displaystyle x}$.

Proof. ${\displaystyle (\Rightarrow )}$ If ${\displaystyle x_{\alpha }\to x}$ and ${\displaystyle V\in {\mathcal {N}}_{x}}$, choose ${\displaystyle \alpha _{0}}$ so that ${\displaystyle x_{\alpha }\in V}$ for all ${\displaystyle \alpha \geq \alpha _{0}}$. Then the tail ${\displaystyle F_{\alpha _{0}}\subseteq V}$, so ${\displaystyle V\in {\mathcal {F}}_{x}}$. Thus ${\displaystyle {\mathcal {N}}_{x}\subseteq {\mathcal {F}}_{x}}$, i.e. ${\displaystyle {\mathcal {F}}_{x}\to x}$.

${\displaystyle (\Leftarrow )}$ If ${\displaystyle {\mathcal {F}}_{x}\to x}$, then for each ${\displaystyle V\in {\mathcal {N}}_{x}}$ we have ${\displaystyle V\in {\mathcal {F}}_{x}}$. Hence some tail ${\displaystyle F_{\alpha _{0}}\subseteq V}$, so the net is eventually in ${\displaystyle V}$. Therefore ${\displaystyle x_{\alpha }\to x}$. ∎

Lemma 3 (cluster points of a net = adherence of its section filter). For a net ${\displaystyle (x_{\alpha })}$,

${\displaystyle \{{\text{cluster points of }}(x_{\alpha })\}\ =\ \operatorname {Adh} {\mathcal {F}}_{x}.}$

Proof. (${\displaystyle \subseteq }$) If ${\displaystyle x}$ is a cluster point, then for every ${\displaystyle \alpha }$ and every neighborhood ${\displaystyle V\ni x}$ there exists ${\displaystyle \beta \geq \alpha }$ with ${\displaystyle x_{\beta }\in V}$. Thus ${\displaystyle V\cap F_{\alpha }\neq \varnothing }$, so ${\displaystyle x\in {\overline {F_{\alpha }}}}$ for all ${\displaystyle \alpha }$. Hence ${\displaystyle x\in \bigcap _{\alpha }{\overline {F_{\alpha }}}\subseteq \operatorname {Adh} {\mathcal {F}}_{x}}$.

(${\displaystyle \supseteq }$) If ${\displaystyle x\in \operatorname {Adh} {\mathcal {F}}_{x}}$, then ${\displaystyle x\in {\overline {F_{\alpha }}}}$ for each ${\displaystyle \alpha }$. Therefore every neighborhood ${\displaystyle V\ni x}$ meets every tail ${\displaystyle F_{\alpha }}$, so for each ${\displaystyle \alpha }$ there exists ${\displaystyle \beta \geq \alpha }$ with ${\displaystyle x_{\beta }\in V}$. This is exactly the cluster-point condition. ∎

Corollary (subnet existence). A point is a cluster point of a net iff it is the limit of some subnet (combine the lemma with the filter theorem “adherent point ⇔ convergent subfilter” and Lemma 1).

Lemma 4 (section filter of a filter-generated net = the filter). Let ${\displaystyle {\mathcal {F}}}$ be a filter on ${\displaystyle X}$ and ${\displaystyle x_{(A,a)}=a}$ its generated net. Then the section filter of this net equals ${\displaystyle {\mathcal {F}}}$.

Proof. Fix ${\displaystyle (A,a)}$. A tail of the net at ${\displaystyle (A,a)}$ consists of all ${\displaystyle x_{(B,b)}=b}$ with ${\displaystyle (A,a)\leq (B,b)}$, i.e. ${\displaystyle B\subseteq A}$. Hence the value set of this tail is

${\displaystyle F_{(A,a)}=\{b:\ \exists B\in {\mathcal {F}},\ B\subseteq A,\ b\in B\}=A.}$

Thus the tail base is precisely ${\displaystyle \{A:\ A\in {\mathcal {F}}\}}$, whose generated filter is ${\displaystyle {\mathcal {F}}}$. ∎

Consequences.

1. ${\displaystyle \operatorname {Adh} {\mathcal {F}}}$ equals the set of cluster points of its generated net.
2. ${\displaystyle {\mathcal {F}}\to x}$ iff the generated net converges to ${\displaystyle x}$ (use Lemma 2).

• Subnets vs. subfilters. Subnet ${\displaystyle \Rightarrow }$ section filter becomes finer (Lemma 1).
• Convergence. A net converges to ${\displaystyle x}$ iff its section filter converges to ${\displaystyle x}$ (Lemma 2).
• Cluster/adherence. Cluster points of a net are exactly the adherence points of its section filter (Lemma 3); dually, adherence points of a filter are cluster points of its generated net (Lemma 4).
• Round-trip. Starting with a filter ${\displaystyle {\mathcal {F}}}$, its generated net’s section filter is ${\displaystyle {\mathcal {F}}}$ itself (Lemma 4).

1. Give a rigorous description of the following sequences of natural numbers:
   (i) ${\displaystyle 1,2,3,4,5\dots }$
   (ii) ${\displaystyle 2,-4,6,-8,10,\ldots }$
   
2. Let ${\displaystyle X}$ be a set and let ${\displaystyle {\mathcal {T}}}$ be a topology over ${\displaystyle X}$. Let ${\displaystyle x\in X}$ and let ${\displaystyle U}$ be a neighbourhood of ${\displaystyle x}$.
   Let ${\displaystyle U_{1}\subset U}$ and ${\displaystyle x\in U_{1}}$. Similarly construct neighbourhoods ${\displaystyle U_{i}\subset U_{i-1}}$ with ${\displaystyle x\in U_{i}\forall i}$. Let ${\displaystyle \left\langle x_{i}\right\rangle }$ be a sequence such that each ${\displaystyle x_{i}\in U_{i}}$.
   
   Prove that ${\displaystyle \lim _{n\to \infty }x_{n}=x}$