I'm hoping someone could review my proof for correctness, thanks!


Show that every metrizable space with a countable dense subset has a countable basis.


Let X be a metrizable space where the metric $d$ induces the topology on X. Let A be a countable subset of X that is dense in X. Then $\bar{A}$ = X.

Then we know that the set of all open balls B$_u$ = { B(x,$\epsilon$)| x$\in$ X, $\epsilon$ > 0 } form a basis for the topology on X.

Consider the following proposed basis:

C$_{a_1}$ = {B(a$_1$,$\epsilon$) | $\epsilon$ = $d$(a$_1$,b) where b $\in$ A for b $\ne$ a$_1$ }

C$_{a_2}$ = {B(a$_2$,$\epsilon$) | $\epsilon$ = $d$(a$_2$,b) where b $\in$ A for b $\ne$ a$_2$ }


C$_{a_n}$ = ...

Now the above set is for every a$_i$ $\in$ A. Hence there are a countable number of sets C$_i$. Now each set C$_i$ has a countable number of elements so we have a countable union of countable sets which is itself countable.

Then have $C$ = $\bigcup$ C$_i$.

Now suppose we have some open set U. Then for every x $\in$ U there is a basis element B(x,$\epsilon$) containing x and B(x,$\epsilon$) $\subseteq$ U. Then we also have B(x,$\epsilon$/2) $\subseteq$ U.

We consider two cases:

If x $\notin$ A then x is a limit point of A since A is dense. Then there is some $a_1$ $\in$ B(x,$\epsilon$/2). Then set the distance H = d(x, a$_1$). Now consider the open ball B(x, H/2). Then some a$_2$ $\ne$ $a_1$ must be in B(x, H/2). Then the set B(a$_2$,r), where r = d(a$_2$,a$_1$), contains x since d(a$_2$,x) $\lt$ H/2 and r $\ge$ H/2. Note: B(a$_2$,r) is a member of C.

Below I show why r $\ge$ H/2:

$$ \text { Suppose r < H/2, then we have d(a$_1$,x) $\leq$ d(a$_1$,$a_2$) + d(a$_2$,x) $\le$ r + H/2. } $$

$$ \text { But d(a$_1$,x) = H so we have: H $\le$ r + H/2 } $$

$$ \text { But we have r < H/2 which leads to the RHS being less than H, a contradiction. Hence r $\ge$ H/2. } $$

$$ \text { So x is contained in the set B(a$_2$,r). } $$

Next we show that for any y $\in$ B(a$_2$,r) we have y$\in$ B(x,$\epsilon$).

We have: d(x,y) $\le$ d(x,a$_2$) + d(a$_2$,y) $\le$ H/2 + r

And: H/2 + r $\le$ H/2 + (3/2)H = 2H since:

r = d(a$_2$,a$_1$) $\le$ d(a$_2$,x) + d(x,a$_1$) $\le$ H/2 + H = (3/2)H.

Now H $\lt$ $\epsilon$/2 since a$_1$ was distance H from x within the B(x, $\epsilon$/2) open ball which contained a$_1$.

Then we have 2H $\lt$ $\epsilon$ which implies y $\in$ B(x,$\epsilon$). Hence we have shown that a set in our proposed basis C contains x and is fully contained within our epsilon neighborhood of x. However, we only considered the case where x is not an element of A.

The case where x $\in$ A can be approached in two ways. If x is a limit point of A we can follow the same idea as above.

If x is not a limit point of A then we show that x must be an isolated point i.e the set {x} is open. Consider if {x} was not open and x was not a limit point of A. Then some open set U containing x must have some point y $\notin$ A and no other points in A. But since U is open in a metric space we can find a B(y,$\epsilon$) $\subseteq$ U. This ball must contain points in A since y is a limit point of A, a contradiction since then U contains points in A. Hence if x is not a limit point of A it must be an isolated point. Since there are countably many points of A we can add these isolated point open sets to the set C and still maintain a countable proposed basis.

This final basis has now been confirmed to produce the same open sets that the uncountable basis B$_u$ produces since any point x $\in$ B(x,$\epsilon$) has an open set in C that contains x and is fully contained within B(x,$\epsilon$). The other direction that B$_u$ can generate every element in C is fairly obvious given it is a metric space. Hence X has a countable basis.


I don't see what makes you think that your set $C_{a_1}$ is countable. If the whole space is uncountable, then there are uncountable many elements $b\neq a_1$ and, in general, there will be uncountably many distinct $\varepsilon$'s with $d(a_1,b)=\varepsilon$.

You can simply take a countable dense subset $A=\{a_1,a_2,\ldots\}$ of $X$ and consider the basis$$\left\{B_{a_n}(q)\,\middle|\,n\in\mathbb N\wedge q\in\mathbb Q\right\}.$$That's a countable basis.

| cite | improve this answer | |
  • $\begingroup$ C_a1 is pairing off a_1 with a countable number of items in A. U can look at it as a function A -> (a1, A). The domain here is countable since A is countable. $\endgroup$ – H_1317 Mar 31 '19 at 9:57
  • $\begingroup$ The element b Must be in A. $\endgroup$ – H_1317 Mar 31 '19 at 9:58
  • $\begingroup$ Right. I missed the “$b\in A$” part. $\endgroup$ – José Carlos Santos Mar 31 '19 at 9:59
  • $\begingroup$ Considering that, is the proof correct? $\endgroup$ – H_1317 Mar 31 '19 at 19:00
  • 1
    $\begingroup$ It looks fine to me. $\endgroup$ – José Carlos Santos Mar 31 '19 at 19:15

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.