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.
Best Answer
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 $L$.