How to show a topological space is not metrizable?

Question

Define the topological space by taking the Reals and identifying all numbers divisible by 10 to a single point, call this space $∆$.

How do we show that $∆$ can not have a metric associated to it and that no space it is homeomorphic to is a metric space?

I think the way to proving that this $∆$ is not metrizable is to show it is not Hausdorff under this topology and the only problem points I can think of are the divisible integers -

So I took 10 and 20 and see that they do not have discrete neighbourhoods but they don't need to since they're identified to the same point. I don't know if this space isn't Hausdorff and can't think of any other way to show it is not metrizable.

Answer 1

Some Hints: Let $(S^1_k)_{k\in\mathbb{Z}}$ be a countable family of copies of $S^1$ and let $p_k\in S_k^1$. Let $X$ be the quotient space obtained from the disjoint union $X'=\coprod_{z\in\mathbb{Z}} S_k^1$ by identifying the points $p_i$. Define a map $f:\Delta\to X$ by sending the image of each interval $[10k,10(k+1)]$ in $\Delta$ homeomorphically onto $S_k^1$ in such a way that $f([10k])=p$ where $p$ is the equivalence class of (any) $p_i$. Prove that $f$ is a homeomorphism.

Argue by contradiction: If $\Delta$ where metrizable with a metric $d$, it induces a metric on $X$ by defining $d'(x,y)=d(f^{-1}(x),f^{-1}(y))$ for $x,y\in X$. The collection $(B(p,1/k))_{k\in\mathbb{N}^*}$ (where $B(u,r)$ is the open ball with center $u$ and radius $r>0$) is a countable neighborhood basis for $p$. From here you can easily derive a contradiction (if not, see Proving infinite wedge sum of circles isn't first countable to find the contradiction).

Answer 2

You can show that the new ideintified point $10\mathbb{Z}$ ,the equivalence class of your identified integers (we could also use $\mathbb{Z}$, it's the same result) does not have a countable local base of neighbourhoods.

And as all metric spaces are first countable, this shows the quotient space is not metrisable.