Answer to Question #217676 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that the unit n-cube Iⁿ is a compact subset of Rⁿ

Answer 1

Lemma 1: Let $X×Y$ be a product space, where Y is compact. If $N$ is an open set of $X×Y$ containing the subset $x_0×Y$ of $X×Y$, then $N$ contains some subset $W×Y$ about $x_0×Y$, where $W$ is a neighborhood of $x_0$ in $X$.

Theorem 2: The product of a finite number of compact spaces is compact.

Proof: Let $X$ and $Y$ be compact spaces. Let $\mathscr{A}$ be an open cover of $X×Y$. Given $x_0\in X$, $x_0×Y$ is compact and hence covered by finitely many elements, say $A_1, ..., A_m$ of $\mathscr{A}$. Their union is an open set containing x0​×Y. By lemma 1, this union contains a subset W×Y about x0​×Y, where W is an open set of X. Hence W is covered by finitely many elements A1​,...,Am​ of A. Thus, ∀x∈X, we can choose a neighborhood Wx​ of x such that Wx​×Y can be covered by finitely many elements of A. The collection of all neighborhoods Wx​ is an open cover of X. So by the compactness of X, ∃ a finite subcover {W1​,...,Wk​} of X. The union of W1​×Y,...,Wk​×Y is X×Y. Since each of them is covered by finitely many elements of A, same applies to X×Y.

The set $[0,1]$ is a compact sub

set of $\mathbb{R}$. So by theorem 2, $[0,1]^n$ is a compact subset of $\mathbb{R}^n$. Hence the unit n-cube is a compact subset of Rn.