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Given $(a,b)$, and $(c,d) \cup (h,k)$ subspaces of $\mathbb{R}$ with the absolute value metric induced topology where $c are any real numbers.

We have just started working with homeomorphisms and I am not sure of cut and dry ways to show that these two spaces are not homeomorphic. We haven't discussed connectedness or covering properties as of when the question was stated.

I am guessing that the spaces are not homeomorphic and I assume it has something to do with $(a,b)$ cannot be constructed by the union of disjoint open intervals but clearly $(c,d) \cup (h,k)$ is such a construction. I am not sure what kind of topological differences this would imply as this is fairly new material and as far as I know the ways that we have to disprove two spaces are homeomorphic are kind of limited with what we have been shown so far.

I do see that a bijection can definitely be constructed, I would imagine, between the spaces but I am thinking though that the spaces are not homeomorphic.

Any advice? =/

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    Your intuition is right. The property they do not share is [Connectedness](http://en.wikipedia.org/wiki/Connected_space). It is a standard exercise to show that if two spaces are homeomorphic and one is connected, so is the other. This allows you to claim they are not homeomorphic, since $(c,d)\cup (h,k)$ is not connected.2012-11-10

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The two spaces are indeed not homeomorphic; the basic reason is that $(a,b)$ is connected, and $(c,d)\cup(h,k)$ is not. Here’s a way to use connectedness even if you haven’t defined it.

Note that the function

$f:(c,d)\cup(h,k)\to\Bbb R:x\mapsto\begin{cases}0,&\text{if }x\in(c,d)\\1,&\text{if }x\in(h,k)\end{cases}$

is continuous. If there were a homeomorphism $h:(a,b)\to(c,d)\cup(h,k)$, the map $f\circ h$ would be a continuous function from $(a,b)$ to $\Bbb R$ whose range was precisely the set $\{0,1\}$. Now use the intermediate value theorem from calculus to show that there is no such continuous function from $(a,b)$ to $\Bbb R$.

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    A nice example of showing something without using more than simple definitions. +12012-11-10