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I'm working my way through Munkres, and I'm having difficulty with exercise 3 of section 40 on the Nagata-Smirnov Metrization Theorem.

Many spaces have countable bases; but no $T_1$ space has a locally finite basis unless it is discrete. Prove this fact.

I'm able to prove one direction (discrete $\rightarrow$ locally finite basis) easily, without even using the $T1$ property:

Let $X$ be a discrete space. Clearly the set of all singletons is a basis for $X$. Therefore, for any point $x \in X$, the set $\{x\}$ is a neighborhood of $x$ which intersects only one basis element, namely itself. That is, $X$ has a locally finite basis.

However, the reverse direction (locally finite basis $\rightarrow$ discrete) is proving to be more difficult. My gut tells me that I should make use of theorem 17.9.

Let $X$ be a space satisfying the $T_1$ axiom; let $A$ be a subset of $X$. Then the point $x$ is a limit point of $A$ if and only if every neighborhood of $x$ contains infinitely many points of $A$.

My attempt so far at proving the contrapositive (not discrete $\rightarrow$ no locally finite basis):

Let $X$ be a $T_1$ space which is not discrete. That is, there is at least one point $x \in X$ such that $\{x\}$ is not open. Thus, every neighborhood of $x$ contains points other than $x$, so that $x$ is a limit point of $X - \{x\}$. By theorem 17.9, every neighborhood of $x$ must then contain infinitely many points of $X - \{x\}$.

From here, my best guess is that I should suppose $X$ has a locally finite basis and try to find a contradiction, but I'm unclear on how to go about this. Just because some neighborhood of $x$ must intersect only finitely many basis elements doesn't necessarily mean it can't contain infinitely many points. Is this an appropriate direction for me to take?

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1 Answer 1

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Recall that if a family $\mathcal A$ of subsets of a topological space $X$ is locally-finite, then it is also point-finite; i.e., for each $x \in X$ the set $\{ A \in \mathcal A : x \in A \}$ is finite.

So we will show that a T1-space $X$ with a point-finite base $\mathcal B$ is discrete.

Note that for each $x \in X$ as $\mathcal B_x := \{ V \in \mathcal B : x \in V \}$ is finite, then $W_x := \bigcap \mathcal B_x$ is an open neighborhood of $x$. Given $y \neq x$, by T1-ness (and the fact that $\mathcal B$ is a base for $X$) there must be a $V \in \mathcal B$ such that $x \in V$ and $y \notin V$. But then $V \in \mathcal B_x$, and so $y \notin \bigcap \mathcal B_x = W_x$. Therefore $W_x = \{ x \}$.

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    $\begingroup$ I see! Munkres only briefly mentions point-finiteness in two exercises, so I wasn't able to recall the property. However, it is rather obvious in retrospect. If some neighborhood of $x$ only intersects a finite number of basis sets, then of course there can only be a finite number of such sets which contain $x$. Thank you. $\endgroup$
    – Itserpol
    Dec 16, 2016 at 17:35
  • $\begingroup$ Really nice proof! $\endgroup$
    – RFZ
    Oct 31, 2019 at 0:15

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