Let \(X\) be a topological and \(\net xn{\bn }\) be a sequence in \(X\). Then \(\net xn{\bn }\) converges to \(x\in X\) provided for each nbhd \(U\) of \(x\) there is \(k\in \bn \) such that \(x_{n}\in U\) for each \(n\ge k\).
Let \(X\) be a topological and \(\net xn{\bn }\) be a sequence in \(X\). Then \(x\in X\) is a cluster point of \(\net xn{\bn }\) provided for each nbhd \(U\) of \(x\) and each \(k\in \bn \) there is \(n\ge k\) such that \(x_{n}\in U\).
The point \(x\) is a cluster point of \(\net xn{\bn }\) if and only if for each nbhd \(U\) of \(x\) the set
\[ \set {n\in \bn :x_{n}\in U} \]
is infinite.
Let \(X\) be a metric space and \(A\sub X\). Then \(x\in \ob A\) if and only if there is a sequence \(\net xn{\bn }\) in \(A\) that converges to \(x\).
Proof. Assume that there is a sequence \(\net xn{\bn }\) in \(A\) that converges to \(x\) and \(U\) be a nbhd of \(x\). Then there is \(k\in \bn \) with \(x_{n}\in U\) for every \(n\ge k\). In particular, \(x_{k}\in U\) so \(A\cap U\neq \emp \). If follows that \(x\in \ob A\).
Now assume that \(x\in \ob A\). Let \(U_{n}:=B\of {x,1/n}\) be an open ball for each \(n\in \bn \). Then \(U_{n}\) is a nbhd of \(x\) for each \(n\in \bn \) so there is \(x_{n}\in A\cap U_{n}\). If \(U\) is any nbhd of \(x\), then there is \(k\in \bn \) with \(U_{k}\sub U\). Then \(x_{k}\in U\) for every \(k\ge n\) so \(\net xn{\bn }\) converges to \(x\). □
Let \(X\) and \(Y\) be metric spaces and \(f:X\to Y\). Then \(f\) is continuous if and only if for every sequence \(\net xn{\bn }\) in \(X\) that converges to \(x\in X\), the sequence \(\cur {f\of {x_{n}}}_{n\in \bn }\) converges to \(f\of x\).
Proof. Assume that \(f\) is continuous and \(\net xn{\bn }\) converges to \(x\in X\). Let \(U\) be a nbhd of \(f\of x\) in \(Y\). Then \(f^{-1}\bof U\) is a nbhd of \(x\) in \(X\) so there is \(k\in \bn \) such that \(x_{n}\in f^{-1}\bof U\) for every \(n\ge k\). Then \(f\of {x_{n}}\in U\) for every \(n\ge k\), implying that \(\cur {f\of {x_{n}}}_{n\in \bn }\) converges to \(f\of x\).
Assume that for every sequence \(\net xn{\bn }\) in \(X\) that converges to \(x\in X\), the sequence \(\cur {f\of {x_{n}}}_{n\in \bn }\) converges to \(f\of x\). Let \(A\sub X\) and \(x\in \ob A\). Then there is a sequence \(\net xn{\bn }\) in \(A\) that converges to \(x\) in \(X\). Since \(\cur {f\of {x_{n}}}_{n\in \bn }\) converges to \(f\of x\), it follows that \(f\of x\in \ob {f\bof A}\). Since \(f\bof {\ob A}\sub \ob {f\bof A}\), it follows that \(f\) is continuous. □
If \(\net xn{\bn }\) is a sequence in a set \(X\) and \(\net kn{\bn }\) is a sequence in \(\bn \) with \(k_{1}<k_{2}<\ds \), then \(\net yn{\bn }\) is a subsequence of \(\net xn{\bn }\), where \(y_{n}:=x_{k_{n}}\) for each \(n\in \bn \).
For any topological space and \(A\sub X\). If a sequence \(\net xn{\bn }\) in \(A\) converges to \(x\), then \(x\in \ob A\).
Let \(X\) be metric space and \(\net xn{\bn }\) be a sequence in \(X\). Then \(x\in X\) is a cluster point of \(\net xn{\bn }\) if and only if there is a subsequence \(\net yn{\bn }\) of \(\net xn{\bn }\) that converges to \(x\).
Proof. Assume that \(x\) is a cluster point of \(\net xn{\bn }\). For each \(n\in \bn \), let \(U_{n}:=B\of {x,1/n}\). Let \(k_{1}\in \bn \) be such that \(x_{k_{1}}\in U_{1}\) and for each \(n\in \bn \), let \(k_{n+1}>k_{n}\) be such that \(x_{k_{n+1}}\in U_{n+1}\). If \(y_{n}:=x_{k_{n}}\)for each \(n\in \bn \), then \(\net yn{\bn }\) is a subsequence of \(\net xn{\bn }\) that converges to \(x\).
Assume that there is a subsequence \(\net yn{\bn }\) of \(\net xn{\bn }\) that converges to \(x\). If \(U\) is a nbhd of \(x\), then there is \(k\in \bn \) such that \(y_{n}\in U\) for every \(n\ge k\). Thus the set \(\set {n\in \bn :x_{n}\in U}\) is infinite, so \(x\) is a cluster point of \(\net xn{\bn }\). □
Let \(X:=\br \) be the set of real numbers with the cocountable topology and \(A:=\br \sem \bq \). Then no sequence \(\net xn{\bn }\) in \(A\) converges to \(0\) since
\[ U:=\br \sem \set {x_{n}:n\in \bn } \]
is open and \(0\in U\). However \(0\in \ob A\) as any nbhd \(U\) of \(0\) is cocountable so \(U\cap A\neq \emp \).
Let
\[ X:=\set {\ang {0,0}}\cup \bn \tm \bn \]
with a topology such that \(\set {\ang {m,n}}\) is open for every \(m,n\in \bn \) and \(U\) containing \(\ang {0,0}\) is open when there is \(m_{0}\in \bn \) such that \(\set {\ang {m,n}\nin U:n\in \bn }\) is finite for every \(m\ge m_{0}\). If \(\net xn{\bn }\) is a bijective sequence in \(\bn \tm \bn \), then \(\ang {0,0}\) is a cluster point of \(\net xn{\bn }\), but no sequence in \(\bn \tm \bn \) converges to \(\ang {0,0}\) so, in particular, no subsequence of \(\net xn{\bn }\) converges to \(\ang {0,0}\).