I was reading the discussion here but I don't find a satisfactory answer. The question is:

Let X = $\mathbb{R}$ with the usual metric and let X′ be a discrete metric space. Describe all continuous functions from X to X′.

Specifically, here's my question, since every set in the discrete metric space is open and closed, then the inverse image of any $U \subset X$ has to be open and closed. I can't find any subsets that are open and closed in $\mathbb{R}$, except for $\mathbb{R}$ itself. Does that mean the only function is $f(x) = k\ \forall x \in \mathbb{R}$? That is, the constant function?


How should I think about it from the connectedness angle? One of the hints given is

discrete metric is totally disconnected. but $\mathbb{R}$ is connected.

I'm not sure how I can use the hint?

  • 2
    $\begingroup$ Yes, that's exactly what it means. As a minor point, you would probably want to say "the constant functions" plural. $\endgroup$ – rnrstopstraffic Nov 16 '17 at 6:40
  • $\begingroup$ @rnrstopstraffic Thank you. How should I think about it from the connectedness angle? $\endgroup$ – user1691278 Nov 16 '17 at 7:20
  • $\begingroup$ @rnrstopstraffic updated the question $\endgroup$ – user1691278 Nov 16 '17 at 7:21
  • $\begingroup$ As you've seen, John Griffin's answer answers your question. It's worth noting that your observation that every set in the descrete space is clopen and that the only clopen sets in $\mathbb{R}$ are $\mathbb{R}$ itself and $\emptyset$ are exactly what it means for $\mathbb{R}$ to be connected and for the discrete space to be (totally) disconnected. $\endgroup$ – rnrstopstraffic Nov 16 '17 at 23:54

The continuous image of a connected space is connected. Since a discrete space is totally disconnected, the only connected subspaces of $X'$ are singletons. Thus the image of a continuous function from $\mathbb{R}$ into $X'$ must be a singleton; i.e., the function must be constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.