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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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1 Answer 1

up vote 4 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).

As a hint, note that finite intersections of open neighbourhoods of $x$ are open neighbourhoods of $x$.

Complete proof. (You've been warned...)

Since $X$ is first-countable, there is a countable local base $\mathcal{B}_x = \{ B_i : i \in \mathbb{N} \}$ at $x$. For each $i \in \mathbb{N}$ we define $$S_i = \bigcap_{j=1}^i B_j.$$ Note for each $i \in \mathbb{N}$ that

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

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 $x$. Given any open neighbourhood $V$ of $x$, as $\mathcal{B}_x$ is a local base at $x$ there is an $i \in \mathbb{N}$ such that $B_i \subseteq V$, and clearly by definition we have $S_i \subseteq B_i \subseteq V$. Thus $\{ S_i : i \in \mathbb{N} \}$ is a local base at $x$. $\dashv$

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I have been thinking about it, and am still not seeing it. Is there a further hint that you could give me? Thanks. –  Alex Petzke Aug 3 '12 at 1:42
@Alex: First note that you can enumerate your original countable basis $\mathcal{B}_x$ at $x$ as $\{ B_n : n = 1 , 2 , \ldots \}$. Next note the following fact: Suppose $\mathcal{B}_x$ is a local basis at $x$ in some topological space, and let $\mathcal{D}_x$ be a family of open neighbourhoods of $x$ such that for each $B \in \mathcal{B}_x$ there is a $S \in \mathcal{D}_x$ such that $S \subseteq B$; then $\mathcal{D}_x$ is also a local basis at $x$. (cont...) –  Arthur Fischer Aug 3 '12 at 2:13
@Alex: (...inued) The trick is then to ensure two things: (1) that $S_{n+1} \subseteq S_n$ for all $n$, and (2) that for each $m$ there is an $n$ such that $S_n \subseteq B_m$. How would intersections help? –  Arthur Fischer Aug 3 '12 at 2:14
If $\{A_n\}_{n=1}^\infty$ is our local nested basis, then we want all the elements of $A_1$ to be in every open neighborhood of $x$. We could do this by taking the intersection of the $S \in \mathcal{D}_x$, since $S \subseteq B_n$ for some $n$. But this isn't (necessarily) a finite intersection. I'm afraid I'm not seeing this. –  Alex Petzke Aug 5 '12 at 17:07
@Alex: I'm sorry but I really do not understand what you are asking. Are you still having trouble with the original problem? or is this something different? (Note that it is generally not possible for a single open neighbourhood of a point to be a subset of all open neighbourhoods of a point; consider the real line under the usual topology.) –  Arthur Fischer Aug 5 '12 at 17:48

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