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Consider $\mathbb{R}$ with the upper limit topology (open sets are of the form $(a,b]$) and consider the subsets $(0,1]$ and $(0, +\infty)$ with the corresponding relative topologies. Show that $(0,1]$ and $(0, +\infty)$ are homeomorphic.

Santiago Vargas Rendón
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HINT: Each of the subspaces is homeomorphic to the Cartesian product $(0,1]\times\Bbb N$, where $\Bbb N$ has the discrete topology. Equivalently, each is the discrete union of countably infinitely many clopen copies of $(0,1]$. This is easy to see for the unbounded subspace, and

$$(0,1]=\bigcup_{n\in\Bbb Z^+}\left(?,\frac1n\right]\;.$$

Brian M. Scott
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  • I'm sorry, what do you mean by "discrete union of countably infinitely clopen copies of $(0,1]$"? – Santiago Vargas Rendón Sep 29 '13 at 21:09
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    @Santiago: Sorry: I accidentally left out the word many. Discrete union and clopen are actually redundant here. The whole thing just means that you have a space that is the union of countably infinitely many pairwise disjoint copies of $(0,1]$ with the upper limit topology, and each of these copies is both an open and a closed set in the union. – Brian M. Scott Sep 29 '13 at 21:11
  • So far this is what I have: First note $$(0,\infty)=\bigcup_{n\in\mathbb{Z}^+} (n-1,n]$$ and $$(0,1]=\bigcup_{n\in\mathbb{Z}^+} \left(\frac{1}{n+1},\frac{1}{n}\right]$$ Then define a function $$f_n=(n-1,n] \longrightarrow \left(\frac{1}{n+1},\frac{1}{n}\right]$$ And then use the pasting lemma. The problem is that I have no idea how to define each function $f_n$ so that it is bicontinuous and bijective. – Santiago Vargas Rendón Sep 30 '13 at 02:01
  • @Santiago: That should be $(n-1,n]$ in the first and third displayed lines, with right-closed intervals, not $(n-1,n)$ with open intervals. Having that right should make it easier. It may also help to replace $f_n$ by a composition of a homeomorphism $g_n:(n-1,n]\to(0,1]$ and a homeomorphism $g_n:(0,1]\to\left(\frac1{n+1},\frac1n\right]$, since those are a bit easier to write down directly. – Brian M. Scott Sep 30 '13 at 02:05
  • Sorry for that mistake, I already corrected it. Also... Aren't those $g_n$'s different? I mean $g_n:(n-1,n]\rightarrow (0,1]$ and $h_n:(0,1]\rightarrow \left(\frac{1}{n+1},\frac{1}{n}\right]$; and then make $f_n=h_n\circ g_n$. – Santiago Vargas Rendón Sep 30 '13 at 02:24
  • @Santiago: Oops; wrong hand triggered. Yes, I meant the second one to be $h_n$, exactly as you have it. – Brian M. Scott Sep 30 '13 at 02:34
  • @Santiago: Let’s see; $g_n$ is certainly right, but $h_n$ doesn’t quite work, because it sends $0$ to $\frac{n}{n+1}$ instead of to $\frac1{n+1}$. The length of the target interval is $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$, and $\frac{x}{n(n+1)}$ goes from $0$ to $\frac1{n(n+1)}$ as $x$ goes from $0$ to $1$, so you want $$h_n(x)=\frac{x}{n(n+1)}+;?$$ – Brian M. Scott Sep 30 '13 at 03:47
  • Sorry for deleting my comment, I immediately realized that $h_n$ was wrong after I added the comment. I think this is the right answer: $$h_n(x)=\frac{x+n}{n(n+1)}$$. – Santiago Vargas Rendón Sep 30 '13 at 04:24
  • @Santiago: Yes, that works fine. – Brian M. Scott Sep 30 '13 at 04:25