I have just learned how to prove that two subspaces of a metric space are topologically equivalent. Now I am trying to learn how to prove that two subspaces are not topologically equivalent.

I have no idea how to go about doing this. Can somebody please show me how to do that for the following simple problem, so that I may apply the same method to more complicated ones?

Prove that on the real line, the intervals $[0,1]$ and $[0,1]\cup[2,3]$ are not topologically equivalent.

I would like to prove this using my book's definition of topological equivalence:

Two metric spaces $(A,d_A)$ and $(B, d_B)$ are said to be topologically equivalent if there are inverse functions $f:A\to B$ and $g: B\to A$ such that $f$ and $g$ are continuous.

So far I have learned to prove continuity using the $\delta$, $\epsilon$ definition of continuity, and I have learned about neighborhoods, limits, and open and closed sets.


  • 1
    $\begingroup$ What properties of continuous function have you learned? Have you learned compactness? etc. $\endgroup$ – BigbearZzz Jul 31 '17 at 14:58
  • 1
    $\begingroup$ It would also be useful to mention the name of the book you are referring. $\endgroup$ – Sahiba Arora Jul 31 '17 at 14:59
  • $\begingroup$ @SahibaArora "Introduction to Topology" by B. Mendelson. $\endgroup$ – Frpzzd Jul 31 '17 at 15:07
  • $\begingroup$ @Nilknarf You haven't answered BigbearZzz's question. $\endgroup$ – Sahiba Arora Jul 31 '17 at 15:17
  • $\begingroup$ @Nilknarf Do you know intermediate value theorem? If so, Carsten's answer should be sufficient. $\endgroup$ – Sahiba Arora Jul 31 '17 at 15:25

Assume that you have a continuous surjective map from $[0,1]$ to $[0,1]\cup[2,3]$ and use the intermediate value theorem to derive a contradiction from that.

(This is of course the same idea as in the earlier answer.)


A continuous image of a connect set is connected.

$[0,1]$ is a connected subset of $\mathbb R$, however $[0,1] \cup [2,3]$ is not.

  • $\begingroup$ Any chance you could prove that they are not equivalent without using connectedness? I haven't learned connectedness in my topology book yet... $\endgroup$ – Frpzzd Jul 31 '17 at 14:54
  • 3
    $\begingroup$ @Nilknarf What have you learned so far then? $\endgroup$ – BigbearZzz Jul 31 '17 at 14:55
  • $\begingroup$ I'll add what I have learned into the question. $\endgroup$ – Frpzzd Jul 31 '17 at 14:55

One elementary topological property that $[0, 1]$ has which $[0, 1] \cup [2, 3]$ doesn't is that the former is connected, whereas the other is not. Showing that an interval is connected is a bit involved, so I'll show that $[0, 1] \cup [2, 3]$ lacks a stronger condition called path-connectedness. We call a space $X$ path-connected if for any two points $a, b \in X$, there exists a continuous function $f: [0, 1] \to X$ such that $f(0) = a, f(1) = b$, which we call a path from $a$ to $b$. Clearly $[0, 1]$ is path-connected, but $[0, 1] \cup [2, 3]$ is not. In particular, you can't make a path from $1$ to $2$. This can be shown by the intermediate value theorem, as the IVT says any path from $1$ to $2$ would also have to go through, for instance, $1.5$.

EDIT: The usual way to show $X$ and $Y$ are topologically distinct is to find some property $P$ that is preserved by topological equivalence, i.e. such that if $A$ has property $P$, and $B$ is topologically equivalent to $A$, then $B$ has property $P$. These are called topological properties. Then show that $X$ has property $P$, but $Y$ does not. In this case, I said that $[0, 1]$ was path-connected, but $[0, 1] \cup [2, 3]$ was not. You should take it as an exercise to prove that if $A$ is path-connected and $B$ is topologically equivalent to $A$, then $B$ is path-connected.


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.