Is the following a valid proof that $X$, an uncountable set, with the cofinite topology is not first-countable?

By way of contradiction, suppose that $\mathscr{B}_x=\left\{B_n:n\in\mathbb{N}\right\}$ is a countable local base of open sets at the point $x$.

Since $X$ is $T_1$ and $\mathscr{B}_x$ is a local base at $x$,

$$\bigcap_{n\in\mathbb{N}} B_n = \{x\}.$$

By De Morgan,

$$X\setminus\bigcap_{n\in\mathbb{N}} B_n = X\setminus\{x\}$$

$$\bigcup_{n\in\mathbb{N}} X\setminus B_n = X\setminus\{x\}.$$

As $B_n$ is open the set $X\setminus B_n$ is finite, hence, it is countable. Notice, we have a countable union of countable sets, hence, a countable set. Therefore, it must be that $X\setminus\{x\}$ is countable, which implies $X$ is countable, a contradiction.

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    $\begingroup$ Looks good to me. $\endgroup$ – triple_sec Jan 31 at 3:39
  • $\begingroup$ Can you argue why $\bigcap\limits_{n\in\mathbb{N}}B_n = \{x\}$ ? $\endgroup$ – jMdA Jan 31 at 3:47
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    $\begingroup$ @jMdA: Yes. Are you asking for the proof? It followings from the fact that $X$ is $T_1$. $\endgroup$ – user853982 Jan 31 at 4:23
  • $\begingroup$ @jMdA. If $x\ne y\in X$ and if $B_x$ is any local base at $x$ then $X\setminus \{y\}$ is a nbhd of $x$ so there exists $b\in B_x$ with $b\subset X\setminus \{y\}.$ So $y\not \in \cap B_x.$ $\endgroup$ – DanielWainfleet Jan 31 at 12:14

Using $\bigcap_n B_n = \{x\}$ is not directly necessary, but is OK in your proof. So your solution is fine as soon as you can justify this intersection from $T_1$-ness (some earlier result in your text presumably).

You could also argue: $\bigcup_n X\setminus B_n$ is at most countable, so we can pick $y \in X$ that is not in this union and unqual to $x$, as $X\setminus \{x\}$ is uncountable. Then $U= X\setminus \{y\}$ is open (definition of cofinite topology, it has finite complement) contains $x$ and so must contain some $B_m$. But then $y$ was chosen to be not in $X\setminus B_m$, so $y \in B_m$ and we cannot have $B_m \subseteq X\setminus \{y\}$, contradiction.

  • $\begingroup$ Took me a second to read through it, but I like this alternate proof as well. Thanks! $\endgroup$ – user853982 Jan 31 at 8:18
  • $\begingroup$ @D.Math Glad you like it. It's slightly more direct as well. $\endgroup$ – Henno Brandsma Jan 31 at 9:07

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