Prove we cannot make a homeomorphism from [0, 1] to ℝ by contradiction.

Question

So start off contradiction proofs by assuming the opposite. So we assume f is a homeomorphism from ℝ to [0, 1].

Since f is a surjection, there exists some ∈ ℝ with $f() = 0$. Let $x_1 = – 1$ and $x_2 = + 1$.

Since f is an injection, and since $x_1≠ x_2 ≠ $,we know that $f(x_1) ≠ f(x_2) ≠ 0$.

Let $M = min \{ f(x_1), f(x_2) \}$.

I know I have to use the intermediate value theorem at some point but I'm stuck at the next step.

Answer 1

Homeomorphisms are continuous. Continuous functions map compact sets to compact sets. $[0,1]$ is compact. $\mathbb{R}$ is not compact. Thus there is no continuous function from $[0,1]$ to $\mathbb{R}$ that has image all of $\mathbb{R}$.

Answer 2

Suppose we have such a homeomorphism. By Heine-Borel, the compact subsets of $\mathbb R $ are those that are closed and bounded. Thus $[0,1] $ is compact and $\mathbb R $ isn't (unbounded). Contradiction.

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One can use connectedness (IVT) to prove that there's no continuous injection from $\mathbb R $ to $I $. Or going in the other direction, one can just remove an endpoint to get that there's no continuous surjection.

One can also argue from compactness that there's no continuous surjection from $I $ to $\mathbb R $.