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In this proofwiki page, it is written without any justification that $\cap \mathcal{B}_x = \{x\}$ where $\mathcal{B}_x$ is some countable basis of a point $x$ in the uncountable set $S$ endowed with the cofinite topology. Why is this true? Indeed, I just proved that it cannot be true in an uncountable space with the cofinite topology. Is this proof flawed? This steps seems crucial in obtaining a contradiction.

What I am trying to do is prove that if $X$ is an uncountable set with the cofinite topology, then it cannot be first countable. I read this, but I couldn't see how to use Brian Scott's hints. Here is what he wrote:

To prove that $X$ is not first countable, you must show that some point of $X$ does not have a countable local base. All points of $X$ ‘look alike’ in the cofinite topology, so it doesn’t matter what point we pick, so let $x\in X$ be any point. Suppose that $\mathscr{B}=\{B_n:n\in\Bbb N\}$ is a countable local base of open sets at the point $x$, meaning that if $U$ is any open nbhd of $x$, then $x\in B_n\subseteq U$ for some $n\in\Bbb N$. For each $n\in\Bbb N$ let $F_n=X\setminus B_n$: $B_n$ is open, so by definition $F_n$ is finite. Let $F=\{x\}\cup\bigcup_{n\in\Bbb N}F_n$; $F$ is the union of countably many finite sets, so $F$ is countable. $X$ is uncountable, so there is some $y\in X\setminus F$. Let $U=X\setminus\{y\}$.

  • Is $U$ an open nbhd of $x$?
  • Is there any $n\in\Bbb N$ such that $x\in B_n\subseteq U$?

I would answer the first question affirmatively; and to answer the question, which is presumably "no", I would suppose that there is such an $n \in \Bbb{N}$. But I am having trouble identifying the contradiction.

Before I even sought help from google, I tried the following (and several variations on it). Let $x \in X$, and let $y \in X - \{x\}$. Then $X-\{y\}$ is open and therefore there is an open set $U_y$ such that $x \in U_y \subseteq X - \{y\}$. Thus we have a mapping $y \mapsto U_y$. I tried to show this mapping is injective, but I couldn't. If I had succeeded, then I believe this would have shown that if there were any basis at $x$, it would have to be uncountable, which means that $X$ is not first countable. But, as you may have noticed, I abjectly failed in this task.

Is there a way to get any of these to proves to work, especially the one I proposed?

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  • $\begingroup$ Doesn't that statement follow from the contradiction hypothesis? That is, the proof begins by assuming the existence of a countable neighborhood basis, then proceeds to arrive at a contradiction. $\endgroup$ – Xander Henderson Oct 25 '17 at 13:11
  • $\begingroup$ Since no point has a countable neighbourhood basis, it is vacuously true. It's probably just badly expressed. Point is, the cofinite topology is $T_1$, so if $\mathcal{B}_x$ is a neighbourhood basis of $x$, then $\bigcap \mathcal{B}_x = \{x\}$. Since the intersection of countably many neighbourhoods of $x$ is larger than $\{x\}$, it follows that $\mathcal{B}_x$ cannot be countable. $\endgroup$ – Daniel Fischer Oct 25 '17 at 13:12
  • $\begingroup$ @XanderHenderson If $\mathcal{B}$ being a countable basis for $x$ implies $\bigcap \mathcal{B} = \{x\}$, then yes. But I don't see why this implication is true. $\endgroup$ – user193319 Oct 25 '17 at 13:13
  • $\begingroup$ Because in a $T_1$ space, that is the case for a neighbourhood basis. $\endgroup$ – Daniel Fischer Oct 25 '17 at 13:15
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Here is a positive proof, which encapsulates the same basic idea, but framing them in a different way.

Let $U_n$, for $n\in\Bbb N$ be any countable number of distinct open sets such that $x\in U_n$ for all $n$. Then for every $n$, there is some finite set $y_n$ such that $U_n=X\setminus y_n$. Therefore, by DeMorgan's law, $$X\setminus\bigcap U_n=\bigcup y_n\quad\text{and so}\quad \bigcap U_n=X\setminus\bigcup y_n.$$

But now $y_n$ are all finite sets, so we have a countable union of finite sets. Therefore this union is countable. So $X\setminus\bigcup y_n=\bigcap U_n$ is a co-countable set, which cannot be a singleton since $X$ is uncountable. So $\bigcap U_n\neq\{x\}$.

In particular, it is impossible that $x$ has a countable neighborhood basis. $\square$

 

The key point here is that when intersecting a countable number of open sets, the complement is a countable union of finite sets, and therefore cannot be equal to $X$ minus a single point.

Finally, your proof idea is flawed. For any $x\in\Bbb R$ consider $U_y=\Bbb R\setminus\{y\}$ as your open set for all $y\neq x$. These are all open sets such that $x\in U_y$. So there are uncountably many of them, since if $y\neq y'$ we have $y\in U_{y'}\setminus U_y$ and $y'\in U_y\setminus U_{y'}$. But nonetheless, $\Bbb R$ is first countable, and in fact second countable.

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The proof just uses the following general fact:

If $X$ is $T_1$ (or equivalently all sets of the form $\{x\}$ are closed) and $x \in X$ and $\mathcal{B}$ is any local base at $x$, then $\bigcap_{B \in \mathcal{B}} B = \{x\}$.

Proof: Of course $x$ is that intersection, so that's a trivial inclusion. Suppose that $p \in \bigcap_{B \in \mathcal{B}} B$ and that $p \neq x$. Then $\{p\}$ is closed and so $x \in X\setminus\{p\}$ which is open so (as we have a local base) for some $B_p \in \mathcal{B}$ we have $x \in B_p \subseteq X\setminus\{p\}$, so $p \notin B_p$, which contradicts $p \in \bigcap_{B \in \mathcal{B}} B$, so $p$ must equal $x$, which proves the other inclusion.

The cocountable topology is $T_1$ as one point sets are countable so closed.

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