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

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

Problem:

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

Proof:

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$$_2$$ }

$$\vdots$$

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.
• Right. I missed the “$b\in A$” part. – José Carlos Santos Mar 31 '19 at 9:59