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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?

EDIT

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

User
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    Deleting a line in $\mathbb{R}^2$ doesn't necessarily mean deleting a line in $\mathbb{R}^3$. – Ningxin Jul 03 '16 at 06:51
  • @ Qiyu Wen, I didn't get you. Can you please elaborate. – User Jul 03 '16 at 06:53
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    Mike's answer is exactly what I mean. – Ningxin Jul 03 '16 at 06:57
  • 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. – Batominovski Jul 03 '16 at 07:16
  • 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. – Henno Brandsma Jul 03 '16 at 09:15
  • @HennoBrandsma so $\Bbb{R}^2$ minus a point is not simply connected while $\Bbb{R}^3$ minus a point is still simply connected ? – MyMathYourMath Jul 15 '22 at 03:34

3 Answers3

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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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    "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) – label Oct 30 '18 at 21:50
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    @Dominik I would argue using Alexander duality in the one point compactification. But this is bigger machinery than most proofs of invariance of domain. –  Oct 30 '18 at 21:54
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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.

Caleb Stanford
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Hint: Note that for some $x_0$, $\Bbb{R}^2\setminus \{x_0\}$ is no longer simply connected while $\Bbb{R}^3\setminus \{x_0\}$ is still simply connected.

Recall: A topological space is simply connected if it is both path connected and every simple closed curve is homotopic to a point.

This seems to be the case each time you increase your dimension of Euclidian space, the space seems to become “more connected”. And using homology, one can deduce $\Bbb{R}^m\cong \Bbb{R}^n$ iff $m=n$. Or an alternate method would be to appeal to the de Rham cohomology groups since they’re all (homology homotopy cohomology) preserved under homeomorphic topological spaces. Basically it all boils down to violating a topological invariant. This is a common technique when trying to show spaces are not homeo. You make an attempt to violate compactness or connectedness or a separation axiom. I.e., some property which is carried over by a continuous function. In your case it’s simply connectedness which is a topological invariant meaning it’s preserved under continuous functions.

I’m not going to rigorously prove that 3D space minus a point is simply connected. You can sort of see it. If you remove origin from plane then considering your simple closed curve to be the unit circle, you cannot contract it to a point but in three space you can “hop” over the hole since you have an added dimension.

MyMathYourMath
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