What is an explicit homeomorphism between cubes $[0,1]^n$ and $[a,b]^n$

Question

I wish to show that $[a,b]^n$ is compact just using the result that $[0,1]^n$ is compact.

In $\mathbb{R}$, we can construct an explicit homeomorphism between $[0,1]$ and $[a,b]$ via $f(x) = (b-a)x+a$, this allows us to establish any closed interval is compact.

Now I wish to demonstrate that any closed cube is compact. Is there a quick homeomorphism we can establish between $[0,1]^n$ and $[a,b]^n$?

If not, is there a easier way to show that $[a,b]^n$ is compact, knowing that $[0,1]^n$ is compact?

Answer 1

Just use the map $$f(x_1,\ldots,x_n)=\left((b-a)x_1+a,\ldots,(b-a)x_n+a\right).$$ This is a continuous map of $[0,1]^n$ onto $[a,b]^n$ (in fact it is a homeomorphism,but this is unnecessary), and since you know $[0,1]^n$ is compact, it follows that $f([0,1]^n)=[a,b]^n$ is compact.

Answer 2

In general, if $\phi: A \rightarrow B$ is a homeomorphism, then we have a homeomorphism $\Psi:A^n \rightarrow B^n$, where $\Psi( \mathbf{x} ) = (\phi(x_1), \phi(x_2), ..., \phi(x_n))$.

So, if you've already found a homeomorphism $[0, 1] \rightarrow [a, b]$, this fact finishes your proof. On the other hand, it isn't hard to just naively construct a homeomorphism/continuous function $[0, 1]^n \rightarrow [a, b]^n$ as others suggest.

Answer 3

$$\lambda(t_1, \ldots, t_n).(a(1-t_1) + bt_1, \ldots, a(1-t_n)+bt_n) : [0, 1]^n \to [a, b]^n$$