$(0,1]$ is not homeomorphic to $(a,b)$ – Carothers Example 5.6(g)

Question

I'm reading Carothers' Real Analysis, and I'm currently looking at homeomorphisms. The author says "two intervals that look different, are different" – i.e. they are not homeomorphic. The proof is done for the case $(0,1]$ and $(a,b)$, where the first interval is semi-open so we expect that the two are not homeomorphic.

I have some difficulty following the book's argument which I paraphrase here (for convenience) with inline questions.

Proof by contradiction. Suppose $(0,1]$ and $(a,b)$ are homeomorphic. Then by removing $1$ from $(0,1]$ we get $(0,1)$, and by removing its image $f(1) = c$ from $(a,b)$ we have that $(0,1)$ is homeomorphic to $(a,c)\cup (c,b)$.

Wait, why is that? I understand that we are deleting an element each from the domain and codomain, so the bijection remains a bijection. What about the continuity of $f$ and $f^{-1}$ though? We literally threw away an element, how do we know that continuity still holds?

But $(0,1)$ is homeomorphic to $\Bbb R$, so $\Bbb R$ would have to be homeomorphic to $(a,c)\cup (c,b)$ too. So $\Bbb R$ can be written as the disjoint union of two non-trivial open sets, which is impossible.

Two questions:

1. Why is $(0,1)$ homeomorphic to $\Bbb R$? We really need to produce a homeomorphism, that's all I'm asking.
2. How to prove that $\Bbb R$ cannot be written as the disjoint union of two non-trivial open sets? The idea that comes to my mind is – we already know that any open subset of $\Bbb R$ can be written as the disjoint union of unique open maximal intervals. Since $\Bbb R$ is open in $\Bbb R$, we can obviously do the same for it as well, and clearly, our choice of intervals would not be maximal in the event that we could write $\Bbb R$ as the union of two disjoint open intervals. Does this make sense?

Thank you!

Answer 1

0. Continuity still holds when you remove a point, because the restriction of a continuous map is still continuous.

1. A homeomorphism between $(0,1)$ and $\mathbb{R}$ is given by $$x \mapsto \tan \left(-\frac{\pi}{2}+t\pi \right)$$

2. $\mathbb{R}$ is connected, so cannot be written as the disjoint union of two non-trivial open sets.