In a previous post on countable network, it was shown that having a countable network is equivalent to being the continuous image of a separable metric space. Since there is an example of a non-metrizable space with countable netowrk, the continuous image of a separable metric space needs not be a separable metric space. However, the perfect image of a separable metrizable space is separable metrizable. First some definitions. A continuous mapping is a closed mapping if is closed in for any closed set . A continuous surjection is a perfect mapping if is closed and is compact for each .

Let be a perfect mapping where has a countable base . Assume is closed under finite unions. Because is a closed mapping, is closed and is open in for each . We show that is a base for . Let and be open with . For each , choose such that . Since is compact, we can choose that cover . Let , which is in . We have . Thus the topology on can be generated by .

**Update (11/24/2009):**

The proof in the above paragraph is faulty. Thanks to Dave Milovich for pointing this out. Here’s the corrected proof.

Let me first prove a lemma.

**Lemma**. Let be a closed mapping and let be open. Then is open in . Furthermore, .

**Proof of Lemma**. Since is a closed mapping, is closed. We claim that . It is clear that . To show that , let . Then cannot be a subset of . Choose . Then we have . Thus and is open. It is straitforward to verify that .

Now I prove that the perfect image of a separable metric space is a separable metric space. Let be a perfect mapping where has a countable base . Assume is closed under finite unions. We show that is a base for .

Let and be open with . For each , choose such that and . Since is compact, we can choose that cover . Let , which is in . Since , we have . We also have . Thus the topology on can be generated by the countable base .

### Like this:

Like Loading...

*Related*

A closed map need not be an open map, even if it’s a perfect surjection between separable metric spaces. Consider the quotient map on the real line that collapses the interval [2,4] to a single point. The map is perfect, yet some open sets, such as (1,3) and (2,4), are mapped to non-open sets.

Thanks, Dave. I really appreciate your pointing out the error.