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If I am examining two sets $X$ and $Y$, each with the discrete topology, will $X\times Y$ have a discrete topology? My understanding is yes. I believe this because $X\times Y$ is the finite product of discrete spaces. Every point in $X$ is open and every point in $Y$ is open, and every point in $X\times Y$ is open. Thus $X\times Y$ has a discrete topology.

Is this understanding correct?

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Yes. In fact, a product of $A$ discrete spaces, with the product topology, each with more than one point is discrete if and only if $A$ is finite. – David Mitra Dec 27 '11 at 16:00
You haven't really explained your argument. Why is it relevant that $X \times Y$ is the finite product of discrete spaces? Why is every point in $X \times Y$ open? – Chris Eagle Dec 27 '11 at 16:01
So, fundamentally, you are correct that to prove a topology on a set $Z$ is discrete, you only need to prove that singletons are open. Why is this enough? Because every subset of $Z$ is a union of singletons, and so if every singleton is open, then so is every subset of $Z$. But, as @ChrisEagle pointed out, you have to prove that singletons in $X\times Y$ are open - you can't just assert it. It depends on your definition of the product topology, however. – Thomas Andrews Dec 27 '11 at 16:25

Given that the projection functions $p_1:X\times Y\rightarrow X$ and $p_2:X\times Y\rightarrow Y$ are continuous, take $(x,y)\in X\times Y$. Clearly: $$\{(x,y)\} = p_1^{-1}(\{x\}) \cap p_2^{-1}(\{y\})$$

But since the topologies on $X$ and $Y$ are discrete, and $p_1$ and $p_2$ are continuous, this means that $p_1^{-1}(\{x\})$ and $p_2^{-1}(\{y\})$ are open in $X\times Y$.

So the singleton set $\{(x,y)\}$ is the intersection of two open sets, and hence is an open set.

Note, this makes clear why this works for finite products, but not infinite products - a singleton in an infinite product would be the infinite intersection of open sets, which is not guaranteed to be open.

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Yes, this is correct. If you want a formal reasoning, it looks something like this.

Make a basis for the topologies of $X$ and $Y$ by letting each one-element (singleton) subset be a basis element. This would produce the discrete topology. Now every subset of $X\times Y$ containing a single point is a product of two basis elements, and thus a basis element itself, and hence we have discrete topology.

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For clarity, I think you should distinguish between a point and a singleton set - that is, every singleton set of $X\times Y$ is a product of two basis elements. A point of $X\times Y$ is not a subset of $X\times Y$, so a point cannot be open, only a singleton set can be open. – Thomas Andrews Dec 27 '11 at 18:16
@ThomasAndrews This is very true, edited. – Arthur Dec 27 '11 at 23:42

Yes. Finite products of discrete spaces are again discrete. Both $X$ and $Y$ have as a base the singleton sets, and the product topology on $X\times Y$ will have as a base, the product of singleton sets, meaning every point is open and closed, and hence the topology on $X\times Y$ is discrete.

An infinite product of discrete spaces need not be discrete however.

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A basis for the product topology on $X \times Y$ is obtained by taking bases $\left\{ U_\alpha\right\}$ and $\left\{ V_\beta \right\}$ of each space $X$ and $Y$, respectively: then, $\left\{ U_\alpha \times V_\beta\right\}$ is a basis of $X \times Y$. Since with the discrete topology you can take as bases of each space all of the points on each space, a basis for $X\times Y$ is the set of all points $(x,y) \in X \times Y$. So $X \times Y$ has the discrete topology indeed.

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Your only flaw is when you say "and every point in $X\times Y$ is open". This is true, but is what you need to justify. To do this, just say "take a point $(a,b)$ in the product, then $\{a\}$ is open in $X$ and $\{b\} $ is open in $Y$. By definition of the product topology, $\{a\}\times \{b\}=\{(a,b)\}$ is open in $X\times Y$."

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