Is $\prod_{n=1}^{\infty} A_{n}$ with product topology, where $A_{n}$=${\{0,1\}}$ has discrete topology for , $n = 1,2,3,\cdots.$ a compact set? How to show it?
My approach: To show that a set is compact, we need to show that every open cover has finite subcover. Now with respect to discrete topology open sets are either singletons or whole space. For Example $A_1 \times A_2\times A_3\times \cdots $ can be covered by $(\{0\}\cup \{1\}) \times (\{0\}\cup \{1\})\times \cdots $ which has finite subcover for each of the $A_i$ but there are countably infinite cartesian product, so what can we conclude from this? However each $A_i$ is itself finite and hence compact w.r.t discrete topology. Is my approach right.