In the last post, I discussed how to efficiently change the topology of a space in order to obtain a locally path-connected space without changing the homotopy or (co)homology groups of the space in question. This is a handy thing to have hanging from your tool belt but there are some reasonable concerns that come along with this type of construction.

Say is metrizable. Must the -coreflection also be metrizable? It turns out the answer is yes but that we might lose separability along the way. In this post, we’ll walk through the details which I learned from some unpublished notes of Greg Conner and David Fearnley.

**Theorem:** If is path-connected and metrizable, then there is a metric inducing the topology of such that the identity function is distance non-increasing.

*Proof.* Suppose is a space whose topology is induced by a metric . Define a distance function on as follows: For any path and , let $$\ell_t(\alpha)=d(\alpha(0),\alpha(t))+d(\alpha(t),\alpha(1)).$$

Observe that for any by the triangle inequality.

Now let $$\ell(\alpha)=\sup\{\ell_t(\alpha)|t\in[0,1]\}$$

For points , we define our metric as $$\rho(a,b)=\inf\{\ell(\alpha)|\alpha\text{ is a path from }a\text{ to }b\}$$

Since for any path from to , we get that showing that the identity is non-increasing.

We should still check that is actually a metric which induces the topology of .

Some notation first: If are paths in such that , then denotes the reverse of and denotes the usual concatenation of paths $$ \alpha\cdot\beta(t)=\begin{cases} \alpha(2t) & 0\leq t\leq 1/2\\ \beta(2t-1) & 1/2\leq t\leq 1 \end{cases}$$

Notice that and given , we have

Thus, in general, . Now we can check that is a metric.

- If , then we may take to be the constant path at this point. Then showing . Conversely, if , consider any path from to . Find such that . Then . Since was arbitrary, we have .
- Symmetry is clear since for every path from to , there is a unique reverse path from to with .
- Suppose . Let be any path from to and be any path from to . Then there is a path is a path from to such that . Therefore finishing the proof that is a metric.

*The metric topology induced by is finer than the topology of :* Suppose is an open set in (with the topology induced by ) and is some path component of . Let . Find an -ball such that . We claim that : if , then so there is a path from to such that . Since for all , we conclude that for all . Since has image in , we must have , proving the claim.

*The topology of is finer than the metric topology induced by :* For the other direction, suppose is an -ball with respect to . Pick a point and let . We claim that the path-component of in is contained in . Let be a path in such that . It suffices to check that . Notice that for all Thus showing that . We now have $$ \rho(x,z)\leq\rho(x,y)+\rho(y,z)<(\epsilon-\delta)+\delta=\epsilon$$proving the claim.

**Example:** One thing to be wary of is that can fail to be separable even if is a compact metric space. For instance, let be a Cantor set in . Then we can use the construction of generalized wedges of circles in the previous post to construct the planar set which is a compact metric space (and certainly separable). This is basically a wedge of circles where the circles are parameterized by a Cantor set. But is an uncountable wedge of circles (with a metric topology – not the CW topology – at the joining point) and this is not separable. The general problem here seems to be that there might be open sets of which have uncountably many path-components!

For any given space , will denote the set of path-components of .

**Theorem:** Let be a metric space. Then is separable if and only if is separable and is countable for every open set .

Proof. If is separable, then since the identity function is continuous and surjective, is separable as the continuous image of a separable space. Now pick a countable dense set and let be a non-empty open set in . Now is the set of path-components of . If , then is open in and thus there is a point . This gives a surjection from a subset of onto showing that is countable.

For the converse, if is a separable metric space then it has a countable basis . Furthermore, we assume is countable for every set . Let be the collection of all path-components of the basic open sets. Then is countable. If is the path-component of in an open set of , then there is a such that Now if is the path-component of in , then where . This shows forms a countable basis for the topology of . Since is metrizable (by our above work), it is also separable.