A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.

Prove that every separable metric space has a countable base.

Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x>0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)<{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.

It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)<\varepsilon_x/4$ and since $d(x, d_j)<{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)<{\varepsilon_x}/{4}+{\varepsilon_x}/{4}<{\varepsilon_x}.$$ Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.

By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r<{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.

Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.

Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.

Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.

  • $\begingroup$ Your proof is correct. But you should at some point mention what your candidate for a countable base is and why it's countable. It's pretty clear for somebody who allready knows the result but maybe a little "hidden" to somebody who has no idea about it. $\endgroup$ – Maik Pickl Jun 3 '16 at 11:55

I'd fill in the details of "it's easy to check" steps. They are crucial here (you must use the triangle inequality somewhere in the proof, as the theorem does not hold for non-metric spaces).

Also state what the supposed countable base is, beforehand. Then show it is indeed countable, and then do this proof to show it's actually a base.

The proof itself is correct in essentials. But fill in all the dots.

  • $\begingroup$ Thanks a lot for remarks! I edited my answer. What do you think about my edited version of proof? $\endgroup$ – ZFR Jun 4 '16 at 9:47
  • $\begingroup$ @RFZ the edited version is much better, yes. $\endgroup$ – Henno Brandsma Jun 5 '16 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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