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.

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

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