Let $X$ is an uncountable set and suppose That $T$ is co-countable topology on $X$ in which $U\in T$ if and only if $U={C}^{c}$ for some finite or countably infinite subset $C\subset X$. Fix $x\in X$ and assume that {$U_n$}$_n$ for $n\in \mathbb{N}$ is a countable family of open sets, each of which contains $x$.

1) Prove that if $y\in U$satisfies $y\ne x$, and that if $V=\{{y}\}^{c}$, then $V$ is a neighborhood of $x$.

2) Prove that there are no natural numbers $n\in \mathbb{N}$ for which $U_n \subset V$.

3) Prove that the topological space $(X,T)$ is not a first countable space.

My solution: 1) {$y$} is finite subset of $X$ so $\{{y}\}^{c}$$\in T$ , $\{{y}\}^{c}$=$U$=$V$, so $\{{y}\}^{c}$ is neighborhood of $X$. I would be thankful if correct my mistake for this prove.

2) There is exist $n\in \mathbb{N}$ such that $U_n \subset V$ and $V=\{{y}\}^{c}$, so $U_n \subset$$\{{y}\}^{c}$, so $U_n$ is finite or countably infinite. Unfortunately, I get stuck here and I do not know how I have continue.

3) If we want to show that $(X,T)$ is not first countable, it is enough to show that: there is exist $x\in X$ such that it has not any countable local basisi in $x$. But I do not know How I define this Basis?

  • $\begingroup$ What is $U$? Is it supposed to be $\bigcap_{n \in \mathbb{N}} U_n$? $\endgroup$ – Trevor Wilson Nov 25 '13 at 23:27
  • $\begingroup$ yes exactly, and it is an open set $\endgroup$ – Zeezee Nov 25 '13 at 23:30
  • $\begingroup$ Do you know what a local basis is? It's not quite clear from your question. $\endgroup$ – Kevin Carlson Nov 25 '13 at 23:32
  • $\begingroup$ for Part 3, I want to show there is not any countable basis, I do not know exactly, If you have other suggestion to prove that$(X,T)$ is not first countable basis I appreciate you. $\endgroup$ – Zeezee Nov 25 '13 at 23:35

1) First see that exist $y \in U$ and $y\neq x$

Suppose $\bigcap_{n \in \mathbb{N}}U_{n}=\{ x \}$, then $X-\{ x \}=\bigcup_{n \in \mathbb{N}}X-U_{n}$ so $X$ is countable ($\rightarrow \leftarrow$)

as $x \in V$ and $V^c=\{ y \} $ is countable then $V$ is neighborhood of $x$.

2) If $U_{n} \subset V$ then $\{ y \} \subset X-U_{n} \Rightarrow y \notin U_{n}$ ($\rightarrow \leftarrow$)

3) By 1) and 2) $(X,T)$ is not a first countable space, being that for any countable family of open neighborhood of $x$ exist a neighborhood $V$ such that $U_{n}\nsubseteq V \,\,\,\forall n \in \mathbb{N}$


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