Why $\Bbb{R}$ and $\Bbb{R}^2$ are not homeomorphic.

Question

Can I argue that if there exists a homeomorphism

$$f: \Bbb{R} \rightarrow \Bbb{R}^2$$

Then subtracting a point should preserve connectedness by continuity of $f$, but then $\Bbb{R}$ minus the origin is disconnected while $\Bbb{R}^2$ minus the origin Is still connected. Is this a good enough argument? As connectedness is a topological property. Which I can prove.

"preserve continuity" should be "preserve connectedness" (continuity is a property of functions, not spaces), but other than that you're good! — Noah Schweber, Jul 12 '22 at 00:13
you're right I meant continuity preserves connectedness @NoahSchweber — homosapien, Jul 12 '22 at 00:14
Well, you need some more tools. The way I like is with homology, which sort of generalizes connectedness. — calc ll, Jul 12 '22 at 00:34

score 4 · Accepted Answer · answered Jul 12 '22 at 00:20

I agree with the commenters that your argument is basically right, however there is one subtlety:

We don't know that $f(0) = 0$. But when we remove a point in $\mathbb{R}$ we should make sure we're removing the corresponding point (as given by $f$) in $\mathbb{R}^2$.

So a slightly better argument would be to say that $\mathbb{R} \setminus \{ 0 \}$ is disconnected, whereas $\mathbb{R}^2 \setminus \{ f(0) \}$ is connected.

Obviously, though, your idea still works.

I hope this helps ^_^

I agree, the later argument you meant is what I had in mind when saying that when you restrict continuity is preserved and thus so should connectedness. — homosapien, Jul 12 '22 at 00:22

score 0 · Answer 2 · answered Jan 15 '23 at 22:24

Let $f : \mathbb{R}^2 → \mathbb{R}$ is a homeomorphism. Then, restricting the domain to $\mathbb{R}^2−\{(0,0)\}$ gives a homeomorphism of the punctured euclidean space to $\mathbb{R}− {f(\{(0,0)\})}$. However, the punctured euclidean space is connected, but as $\mathbb{R}− {f(\{(0,0)\})}$ is not connected, because note that $$\mathbb{R}− {f(\{(0,0)\})} = (−\infty, f(\{(0,0)\})) \cup (f(\{(0,0)\}),\infty)$$ so the open sets $(−\infty, f(\{(0,0)\}))$ and $(f(\{(0,0)\}),\infty)$ give a separation of this space. Hence, the punctured euclidean space and $\mathbb{R}− {f(\{(0,0)\})}$ are not homeomorphic, a contradiction. Therefore, we conclude that $\mathbb{R}^2$ and $\mathbb{R}$ are not homeomorphic.

What does this add to the existing (accepted) answer? – Noah Schweber Jan 15 '23 at 22:39 — Noah Schweber, Jan 15 '23 at 22:39

Why $\Bbb{R}$ and $\Bbb{R}^2$ are not homeomorphic.

2 Answers2