Topologically, what distinguishes $\mathbb{R}$ and $\mathbb{R}^2$

Question

I thought this question was easy, but turns out to be a little bit tricky

Let us give $\mathbb{R}$ and $\mathbb{R}^2$ the usual topology. Then, $\mathbb{R}$ and $\mathbb{R}^2$ are:

1. both uncountable
2. both separable, first countable, second countable, suslin
3. metrizable, not compact
4. connected

$$\vdots$$ Seems we require a new concept:

Let $x \in X$ be a cut point if $X \backslash \{x\}$ has a non-trivial clopen subset.

We know this is a topological invariant. See my answer here: Show that cuts are preserved under homeomorphism

It seems that the only distinguishing feature is that $\mathbb{R}$ have different amount of cut points than $\mathbb{R}^2$.

But how many cut points?

Obviously it only takes one to cut $\mathbb{R}$, and it takes uncountable many (like a line) to cut $\mathbb{R}^2$.

And how do we know whether taking away uncountable many points from $\mathbb{R}$ doesn't also produce a clopen subset? In that case, $\mathbb{R}$ and $\mathbb{R}^2$ would have same amount of cut points, thus not topologically distinguishable.

Can someone resolve this for me?

Answer 1

The point is that every point of $\mathbb{R}$ is a cut point, while no point of $\mathbb{R}^2$ is; together with the fact that if $f:X\cong Y$ is a homeomorphism and $x$ is a cut point of $X$, then $f(x)$ is a cut point of $Y$. With these two facts in hand, it's clear that $\mathbb{R}\not\cong\mathbb{R}^2$: if $f:\mathbb{R}\cong\mathbb{R}^2$, then $0$ is a cut point of $\mathbb{R}$ but $f(0)$ (whatever that is) isn't a cut point of $\mathbb{R}^2$.

I don't quite understand your second-to-last paragraph, about removing many points at once; can you elaborate?

Answer 2

If you're in the market for more distinguishing features, here's another property that should be preserved under homeomorphism:

In $\mathbb{R}$, a subset $S$ is connected $\implies S$ is also path connected. In $\mathbb{R}^2$, however, this is not the case. See, e.g. the topologist's sine curve.