I was reading Munkres for Topology. In that, it is mentioned that $\mathbb R$ is not homeomorphic to $\mathbb R^2$ as deleting a point from both makes the first one disconnected while the latter one still remains connected.

Can't we say on the same lines that $\mathbb R^2$ is not homeomorphic to $\mathbb R^3$ as deleting a line from both makes the first one disconnected but second one is still connected.

Where I'm going wrong?


I know we can show them to be non homeomorphic using simple connectedness, but I want to find the error in that approach.

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    $\begingroup$ Deleting a line in $\mathbb{R}^2$ doesn't necessarily mean deleting a line in $\mathbb{R}^3$. $\endgroup$ – Qiyu Wen Jul 3 '16 at 6:51
  • $\begingroup$ @ Qiyu Wen, I didn't get you. Can you please elaborate. $\endgroup$ – User Jul 3 '16 at 6:53
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    $\begingroup$ Mike's answer is exactly what I mean. $\endgroup$ – Qiyu Wen Jul 3 '16 at 6:57
  • $\begingroup$ The easiest proof I know is using the fact that $\mathbb{R}^2\setminus\{*\}$ and $\mathbb{R}^3\setminus\{*\}$, where $*$ is a point, have different top homology groups. $\endgroup$ – Batominovski Jul 3 '16 at 7:16
  • $\begingroup$ Simple connectedness works for 2 and 3, but beyond that... As a topologist I prefer the Brouwer approach: define a dimension function that is a topological invariant and show $\dim(\mathbb{R}^n) = n$. Munkres also does dimension theory (the beginning of it) in his topology book. $\endgroup$ – Henno Brandsma Jul 3 '16 at 9:15

Key fact: "$x$ is a point in topological space $X$" is a property invariant under homeomorphism, whereas "$x$ is a line in topological space $X$" (whatever definition of line you take) is not invariant under homeomorphism.

The argument that $\mathbb{R}$ and $\mathbb{R}^2$ are not homeomorphic goes through because, supposing $\mathbb{R}$ were somehow homeomorphic to $\mathbb{R}^2$ we know the removed point in $\mathbb{R}$ corresponds to a point in $\mathbb{R}^2$. But not so with a line in $\mathbb{R}^2$ corresponding to a line in $\mathbb{R}^3$.

Of course, you could instead of saying remove a line from $\mathbb{R}^2$, remove a copy of $\mathbb{R}$ (a subspace homeomorphic to $\mathbb{R}$. This is now invariant under homeomorphism, however, you have essentially the same problem: a copy of $\mathbb{R}$ in $\mathbb{R}^3$ need not be anything nice, and certainly need not be a line.

Another way of saying this is that the "line" in $\mathbb{R}^3$ can't be chosen by you; it has to be the image of the line in $\mathbb{R}^2$ under the homeomorphism. To show topological spaces $X$ and $Y$ are not homeomorphic, it certainly is not sufficient to remove the same subspace from both (that you pick) and show that the results are not homeomorphic spaces. A counterexample is deleting the open interval $(0,1)$ from $\mathbb{R}$ and $(0,2)$ (homeomorphic topological spaces); the former becomes disconnected; the latter remains connected.


Let's try to reproduce the argument.

Suppose there was a homeomorphism $f: \Bbb R^2 \to \Bbb R^3$. Call your line $L \subset \Bbb R^2$. Then it would restrict to a homeomorphism $\Bbb R^2\setminus L \to \Bbb R^3 \setminus f(L)$.

Now, why is $\Bbb R^3 \setminus f(L)$ connected? Why can't the line $f(L)$ be embedded in an extremely bizarre way that causes its complement to be disconnected? (As an example of 'strange embeddings', see the Alexander Horned sphere, an embedding of $S^2$ into $\Bbb R^3$ such that one of two connected components of its complement is not simply connected.)

The way you get around this when $L$ is a point instead of a line is that it's hard to embed a point strangely! A point is a point is a point, and we know that $\Bbb R^3 \setminus \{p\}$ is simply connected for any choice of point $p$. But it's hard to see that $\Bbb R^3 \setminus L$ is connected for any choice of properly embedded real line $L$.

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    $\begingroup$ thanks for the explanation. $\endgroup$ – User Jul 3 '16 at 7:00
  • $\begingroup$ "But it's hard to see that $\mathbb{R}^3 ∖ L$ is connected for any choice of properly embedded real line $L$." What is needed to see this? Is it actually possible to finish the proof along these lines? (pun intended) $\endgroup$ – label Oct 30 '18 at 21:50
  • $\begingroup$ @Dominik I would argue using Alexander duality in the one point compactification. But this is bigger machinery than most proofs of invariance of domain. $\endgroup$ – user98602 Oct 30 '18 at 21:54

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