Last time, we learned that—in a world in which much is uncertain—at least we can trust continuous maps of Hausdorff spaces to behave nicely with respect to dense subsets.

Or can we? We showed that if is dense, then any continuous map is determined by its values on , which jives well with our intuition about maps , for example.

But in we can go ever farther. For any open set , we can construct a bump function that is nonzero on , but zero outside of . It follows that if is not dense, then continuous functions are not determined by their values on .

Is this true of all Hausdorff spaces?

The answer is yes, but proving it requires some creativity. The brute force approach does not work here—there is no clear way to create a bump function on some arbitrary space. If you consider yourself a point-set topology guru, I encourage you to try to prove the proposition yourself before reading on.

Proposition: If is a Hausdorff space and is not dense, then there exists some Hausdorff space and two distinct continuous functions that agree on .

Proof: Let , and let be two copies of glued together along . Since there are points not in , the two embeddings are distinct, but agree on by construction.

I have kept the proof (extremely) short to illustrate the beauty of the idea, but there are a few assumptions that need to be justified. The most serious of these is the assertion that is Hausdorff. But a little reading on disjoint unions and quotient spaces will reveal that this is not particularly difficult. (though it is very necessary to know that is closed)

I do not know if this result is as strong as possible. So I leave the question to you, if you hunger and thirst for more than these two articles have provided.

Question: Does there exist some space and a non-dense subset such that any continuous map , where is Hausdorff, is determined by its values on ?

Share this: Twitter

Facebook

Like this: Like Loading... Related