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I want to prove that the map $I \rightarrow I \times I$ where $0.a_{1}a_{2}a_{3}... \rightarrow (0.a_{1}a_{3}a_{5}..., 0.a_{2}a_{4}a_{6}...)$ is not a homeomorphism. Note that $I$ is the unit interval from $0$ to $1. My thought on this problem is that the reason it is not a homeomorphism is the fact that it is not surjective. I am unsure how I would write this as a proof if that is the right way to go. The help would be greatly appreciated.

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A homeomorphism is defined as a bijective continuous map with continuous inverse. –  Mike Miller Feb 23 at 7:18
    
I know. That is why if its not surjective, then its not bijective. –  user128349 Feb 23 at 7:30
    
And thus you've proven that the given map is not a homeomorphism. (However, this map is surjective. It's not injective.) –  Mike Miller Feb 23 at 7:32
    
That makes sense to me. I just wanted to mathematically show it though to prove it ;) –  user128349 Feb 23 at 7:33
    
If your map wasn't surjective, then that would be a proof. –  Mike Miller Feb 23 at 7:33

1 Answer 1

Your map is surjective. You need to show that it's not injective. (It's actually even further afield of being a homeomorphism: it's not even continuous.) So you want some element of $I \times I$ to be mapped to by two different elements of $I$.

Hint: Use that $0.0999... = 0.10000...$

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Okay that makes sense. I see your argument. Quick question though. Are you sure it is surjective? I might be over thinking it but I just want to make sure. You can think of $I$ as the unit real line and $I \times I$ as a unit square if that makes sense. How does every element in the image get hit from some element in the pre image (unit real line)? I didnt think $I$ could cover it all. –  user128349 Feb 23 at 7:51
    
@user128349 Every couple $(0.b_1b_2\dots,0.c_1c_2\dots)$ gets hit by $0.b_1c_1b_2c_2\dots$. –  Pece Feb 23 at 8:57

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