Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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 $x$. $\dashv$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.