# Let $X$ be the two point set with the discrete topology. Let $X^\omega$ be the countable infinite cartesian product of $X$ with itself.

Let $X$ be the two point set with the discrete topology. Let $X^\omega$ be the countable infinite Cartesian product of $X$ with itself.

Prove or disprove: the box topology on $X^\omega$ is discrete.

So what I don't get is what a two point set is? I'm guessing it is the set $\{0,1\}$. To prove something is discrete space we have to show that every subset of $X$ is open (and hence closed) but I don't know how to go about doing this? Please can someone help me with the proof?

The box topology is an infinite product of open sets.

Since the base space is discrete, taking arbitrary products of discrete sets gives discrete sets. An arbitrary element of $X^\omega$ in this topology can be thought of as just ({$x$1},{$x$2},...,) which an infinite product of open sets, and is open in the box topology. This proves that the box topology gives the discrete topology on the infinite product as claimed.

• Please tell us what you've tried. Also, please ask one question at a time. – Shaun Oct 19 '17 at 18:26
• With the box topology, it is discrete. With thr product topology it is homeomorphic to the Cantor set. – William Elliot Oct 19 '17 at 18:38
• I have no idea how to start this proof. Please can someone tell me how to start it... – Anjeli Ford Oct 19 '17 at 18:55
• @AnjeliFord Start with the definition of "discrete". How do you prove/disprove that a space is discrete? – John Griffin Oct 19 '17 at 21:13
• You prove a space is discrete by showing every subset of X is open (and hence closed)... – Anjeli Ford Oct 20 '17 at 2:42