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From this post it seems that proving $\mathbb{R}^n$ isn't homeomorphic to $\mathbb{R}^m$ for $m \neq n$ is not elementary:

It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>1$: subtract a point and use the fact that connectedness is a homeomorphism invariant.

Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>2$ by subtracting a point and checking if the resulting space is simply connected. Still straightforward, but a good deal less elementary.

WLOG we can assume $m<n$. When $m=2$, can I subtract a line and prove the result using the fact that connectedness is a homeomorphism invariant.

Generally can I just subtract a “$m-1$ dimensional line” $\{(a_1, \dots,a_{m-1},0)\mid a_i\in\mathbb R\}$ and done the proof?

I am aware that things cannot be such simple, but I can not find where I’m wrong.

Thanks for your help.

Evan Tseng
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  • When you say "subtract a line," I interpret that as "subtract a subspace which is homeomorphic to a line," for example $(0,1)$ as a subset of $\mathbb{R}$. If $\mathbb{R}$ is embedded into $\mathbb{R}^2$ as the $x$-axis, then we have a "line" $(0,1) \times {0}$, the complement of which is connected. – John Palmieri Jul 24 '23 at 16:56
  • Well, yes, you subtract a line in $\mathbb R^2$ - this corresponds to subtracting the image of that line in $\mathbb R^n$, in the purported homeomorphism map. You need to prove that the image of that line (which may not be a line anymore) cannot split $\mathbb R^n$ into two components. A continuous image of a line inside $\mathbb R^n$ (e.g. $\mathbb R^3$) can be very hairy indeed. –  Jul 24 '23 at 16:58
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    And in fact, you probably want to look at the second most-voted answer in the post you linked to for a much more expanded (and much more beautifully presented) argument that I was trying to put forward above: https://math.stackexchange.com/a/24900/700480 –  Jul 24 '23 at 17:03
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    @StinkingBishop Thanks for your reply. Now I wan’t to ask if the statement “the image of that line (which may not be a line anymore) cannot split Rn into two components” is actually true. – Evan Tseng Jul 25 '23 at 03:13

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