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I am trying to prove the following:

Let $X$ be a first countable space and $x$ a member of $X$. Prove that there is a local nested basis $\{S_n\}_{n=1}^\infty$ at $x$.

Since $X$ is first countable there is a countable local base $\mathcal{B}_x$ at $x$. Constructing a nested sequence of subsets of $\mathcal{B}_x$ is easy. Let $B_1 \in \mathcal{B}_x$. Then $B_1$ is an open set containing $x$, and so contains a member $B_2$ of $\mathcal{B}_x$ by the definition of a local base. Then $B_2$ is an open set containing $x$ and so contains a member $B_3$ of $\mathcal{B}_x$. Continuing in this fashion we obtain a nested sequence $\{B_n\}_{n=1}^\infty$ of members of $\mathcal{B}_x$ containing $x$.

I'm having trouble showing that this is a local base, in that I don't see how to prove that every open set in X contains some member of this sequence. Of course it's possible that this isn't true, and there's a different way to construct the nested sequence so that this can be done, but I'm having trouble with this as well.

Which way do I need to proceed, and how do we complete the proof? Thanks.

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up vote 6 down vote accepted

Note that although you are correct that there is some $B_2 \in \mathcal{B}_x$ such that $B_2 \subseteq B_1$, since $B_1 \in \mathcal{B}_x$ it might be that $B_2 = B_1$, and this could continue ad infinitum. If this were to happen (at any point along the inductive construction) it would almost certainly not generate a basis at $x$ (unless $x$ has a smallest open neighbourhood and we picked it at some point).

To avoid this problem we begin by first fixing any countable local base $\{ B_i \}_{i \in \mathbb{N}}$ at $x$. Next, for each $i \in \mathbb{N}$ we define $$S_i := B_1 \cap \cdots \cap B_i.$$ Note for each $i \in \mathbb{N}$ that

  • as the set $S_i$ is a finite intersection of open neighbourhoods of $x$, it is itself an open neighbourhood of $x$; and
  • $S_{i+1} \subseteq S_{i}.$

Therefore $\{ S_i \}_{i \in \mathbb{N}}$ is a nested family of open neighbourhoods of $x$, so we need only show that it is a local base at $x$. But this follows immediately from the fact that $S_i \subseteq B_i$ for each $i$, and we chose $\{ B_i \}_{i \in \mathbb{N}}$ to be a local base at $x$.

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