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The spaces are $\mathbb{R} \times [0,1]$ and $\mathbb{R} \times (-\infty,0]$.

So, if [0,1] was either [0,1) or (0,1] showing homeomorphism is trivial. However, the inclusion of both the points leads me to believe that there will not even exist a (surjective) continuous map between the two spaces.

Adienl
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    There are plenty of continuous maps (just as there are plenty of continuous maps from $[0,1]$ to $(-\infty,0]$, and plenty in the other direction). Continuous, bijective maps, on the other hand, are more difficult to find. – Arthur Sep 04 '19 at 19:32
  • @Arthur No. If there was then the image of $[0,1]$ would be bounded while the other set is not. – Adienl Sep 04 '19 at 19:34
  • However, even that observation will not really help us with a formal proof for this!! – Adienl Sep 04 '19 at 19:35
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    It sounds like you're restricting attention to surjections. There are plenty of continuous maps from $\mathbb{R}\times[0,1]$ to $\mathbb{R}\times (-\infty, 0]$, such as $$(a,b)\mapsto (a,b-17).$$ – Noah Schweber Sep 04 '19 at 19:37
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    $x\mapsto x-1$ is a continuous map from $[0,1]$ to $(-\infty,0]$. And $x\mapsto -\frac{1}{x-1}$ is a continuous map from $(-\infty,0]$ to $[0,1]$. And there are many more where those came from. – Arthur Sep 04 '19 at 19:37
  • @Arthur Haha! Let's keep the maps surjective. xP – Adienl Sep 04 '19 at 19:39
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    @Adienl Sure. But then you ought to say so from the get-go. Your question post says "there will not even exist a continuous map between the two spaces." That's false, and it was what I was responding to. – Arthur Sep 04 '19 at 19:40
  • @Arthur. Sorry! I will edit that right away. – Adienl Sep 04 '19 at 19:41
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    I think you can show that these spaces are not homeomorphic by using ends. The first space has two ends but the second has only one end. You can see this answer where I use ends as an example. – Adam Chalumeau Sep 04 '19 at 19:47
  • @AdamChalumeau Seems pretty interesting. I shall have a look! – Adienl Sep 04 '19 at 19:53

3 Answers3

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No. The only proof I can come up with uses some algebraic topology:

Both $\mathbb{R}\times[0,1]$ and $\mathbb{R}\times(-\infty,0]$ are topological manifolds with boundaries. The boundary of $\mathbb{R}\times[0,1]$ consists of two lines, and is disconnected. The boundary of $\mathbb{R}\times(-\infty,0]$ is a line, and connected.

However, any homeomorphism between topological manifolds with boundaries restricts to a homeomorphism between their boundaries; see

(the boundary of a manifold is the complement of its interior) so those spaces are not homeomorphic.

Luiz Cordeiro
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    I don't know algebraic topology. I suppose I should start reading up on it. Thanks for the answer! – Adienl Sep 04 '19 at 19:58
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    I guess you don't in fact need all that much algebraic topology to apply your argument; just knowing that the circle is not contractible is enough. For both these spaces, you can tell whether a point is a boundary point or not according to whether its removal leaves you with a contractible space or space that is homotopy equivalent to $S^1$. – Mike F Sep 04 '19 at 20:52
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I think Luiz's answer is right on the money, but here is an alternative anyway. If these spaces were homeomorphic, then so would be their 1-point compactifications. However, the 1-point compactification of $\mathbb{R} \times [0,1]$ is homotopy equivalent to the circle $S^1$, whereas the 1-point compactification of $\mathbb{R} \times (-\infty,0]$ is contractible (it is homeomorphic to the closed 2-disk). So, as long as you have enough tools to know that the circle is not contractible, you can apply these observations to conclude that the spaces are not homeomorphic.

Mike F
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Here's an approach that doesn't use any algebraic topology; you only need Heine-Borel. If you remove a compact subset from $\mathbb{R} \times (-\infty,0]$ the remainder has only one (connected) component whose closure is not compact. If you do the same to $\mathbb{R} \times [0,1]$ the remainder can have two such components.

David Hartley
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  • This does not work; homeomorphisms between open subsets of $\mathbb{R}^2$ do not necessarily extend to their closures. Alternatively, change $\mathbb{R}\times(-\infty,0]$ and $\mathbb{R}\times[0,1]$ by the homeomorphic sets $(0,1)\times(0,1]$ and $(0,1)\times[0,1]$, so the argument does not work anymore (all closures of their subsets are compact). – Luiz Cordeiro Sep 09 '19 at 03:12
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    But homoemorphisms between whole spaces do respect the closures of their subsets. A homoemorphism between $X$ and $Y$ will take components of $X\setminus K$ to components of $Y \setminus f(K)$ and their closures (in $X$) to the closures of the components in $Y$. A component with compact closure in $X$ will correspond to one with the same property in $Y$. (Essentially my answer is just an attempt at a proof that the one space has two ends and the other only one, without bothering to define ends. I see now that Adam Chalumeau had already suggested using ends in a comment.) – David Hartley Sep 10 '19 at 07:21