Question #217676

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


1
Expert's answer
2021-07-20T06:45:29-0400

Lemma 1: Let X×YX×Y be a product space, where Y is compact. If NN is an open set of X×YX×Y containing the subset x0×Yx_0×Y of X×YX×Y, then NN contains some subset W×YW×Y about x0×Yx_0×Y, where WW is a neighborhood of x0x_0 in XX.


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


Proof: Let XX and YY be compact spaces. Let A\mathscr{A} be an open cover of X×YX×Y. Given x0Xx_0\in X, x0×Yx_0×Y is compact and hence covered by finitely many elements, say A1,...,AmA_1, ..., A_m of A\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][0,1] is a compact sub

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

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