# Nested interestection forms neighbourhood basis

Let $$X$$ be a topological space and $$x \in X$$. Suppose that there exists a countable collection $$(U_n)_{n \geq 1}$$ of open sets such that $$U_{n + 1} \subseteq U_n$$ and $$\bigcap\limits_{n \geq 1} U_n = \{ x \}$$. Is it true that $$U_n$$ is a neighbourhood basis of $$x$$? This is equivalent to: if we choose $$x_n \in U_n$$, then the sequence $$(x_n)_{n \geq 1}$$ converges to $$x$$.

What if we add some separability axiom, like Hausdorff? What if we add local compactness?

The general answer is "no", even for a very good space like $$\mathbb{R}$$.

Take $$X=\mathbb{R}$$, $$x=0$$ and put

$$U_n=(-1/n, 1/n)\cup (n,\infty)$$

and note that $$\bigcap U_n=\{0\}$$ even though there exists a sequence $$a_n\in U_n$$ not convergent to $$0$$, namely $$a_n=n+1$$.

Also note that obviously $$U_n$$ don't form neighbourhood basis since each one of them is unbounded.

• This does not work entirely as $U_{n+1}\not\subset U_{n}$. – Floris Claassens Feb 12 at 14:40
• @FlorisClaassens Well, the indexing was off. I've simplified it. – freakish Feb 12 at 14:40
• I see, this looks good. – Floris Claassens Feb 12 at 14:42