# Is it compact in product topology?

Define $A_n=\{0,1\}$ for $n=1,2,3,\cdots$ and suppose it is given discrete topology, then will $\prod^{\infty}_{n=1} A_n$ with product topology be compact?

Since $A_n$ is compact, by Tychonoff's theorem the product will be compact. Am I correct?

• Yes, that's correct. – David Mitra Dec 21 '15 at 8:42
• Tychonoff's theorem – user295959 Dec 21 '15 at 8:43

Yes. Being finite, $A_n$ is compact, since for all subsets of $A_n$ each of its open covers is finite (and thus it is its own finite subcover).
Tychonoff's theorem assures compactness of arbitrary products of compact spaces, therefore also $\prod\limits_{n\in\mathbb{N}} A_n$ is compact.
• Good. However, since Tychonoff's theorem depends on the axiom of choice, one may wish to give a direct proof for the space $\prod_{n-1}^\infty A_n$ by showing that it is homeomorphic to the Cantor set, a closed and bounded subset of $\mathbb R.$ – bof Dec 21 '15 at 9:45