# How to *disprove* topological equivalence

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.

Thanks!

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

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