Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
1  
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
add comment

5 Answers

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.

share|improve this answer
1  
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
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

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 $X\times Y$."

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.