# A simple, concrete example of the product topology

I am having some hard time understanding the concept of product topology.

Could somebody give me a simple, concrete example of a product topology? For example, what is the product topology of $A\times B$ when both $A$ and $B$ are equipped with the discrete topology?

Better, let's say that $A$ and $B$ each consist of two points $a_1,a_2$ and $b_1,b_2$ respectively. What is the product topology of $A$ and $B$?

• For a finite product, product topology coincides with box topology. – Sahiba Arora Aug 2 '17 at 14:45
• Do you know that the collection of sets $U\times V$ with $U$ open in $A$ and $V$ open in $B$ is a base of the product topology? So a set is open in the product topology iff you can write the set as a union of this sort of sets. Now find these bases in the cases you mention and have a look at the unions of their elements. – drhab Aug 2 '17 at 14:46
• For better understanding, compare the box and product topology on $\mathbb R^{\omega}.$ – Sahiba Arora Aug 2 '17 at 14:47
• @SahibaArora Regarding the nature of the question it might be that the OP is not familiar yet with box topology. – drhab Aug 2 '17 at 14:49
• The plane, $\Bbb R^2$, with the usual topology, is the product $\Bbb R\times \Bbb R$ of the real line with itself. – MJD Aug 2 '17 at 15:11

If both $A$ and $B$ are equipped with the discrete topology, the the product topology on $A\times B$ is again the discrete topology. This is so because if $a\in A$ and $b\in B$, then $\{(a,b)\}=\{a\}\times\{b\}$ and therefore $\{(a,b)\}$ is the cartesian product of two open sets. So, $\{(a,b)\}$ is an open set. Since all singletons are open sets, the topology is the discrete topology.

• Thanks Jose for the reply. If $A$ and $B$ are both discrete spaces, they each have 4 elements. How many elements are there in the product topology? Is it 16, the simple Cartesian product of both topologies? – user60264 Aug 2 '17 at 14:54
• @student Why do you say that if $A$ is a discrete space, then it has $4$ elements? It can have any number of elements. – José Carlos Santos Aug 2 '17 at 14:56
• because each space has two points – user60264 Aug 2 '17 at 14:57
• @student What I wrote was about any two discrete spaces $A$ and $B$. – José Carlos Santos Aug 2 '17 at 14:58

Let $T_A$ be the topology on $A$ and let $T_B$ be the topology on $B.$ Consider any topology $T$ on $A\times B$ such that the projections $p_A: (A\times B)\to A$ and $p_B:(A\times B)\to B$ are continuous, where $p_A((a,b))=a$ and $p_B((a,b))=b.$

If $A'\in T_A$ and $B'\in T_B$ we must have $A'\times B=p_A^{-1}A'\in T$ and $A\times B'=p_B^{-1}B'\in T,$ so $$T \supset U=\{A'\times B:A'\in T_A\}\cup \{A\times B':B'\in T_B\}.$$ Now $U$ is a sub-base for a topology $T^*$ on $A\times B,$ so $T\supset T^*.$

So any $T$ for which $p_A$ and $p_B$ are continuous is stronger than $T^*.$ And we can confirm that $p_A$ and $p_B$ are continuous when $A\times B$ is given the topology $T^*.$ So $T^*$ is the weakest topology on $A\times B$ such that $p_A$ and $p_B$ are continuous.

$T^*$ is called the product topology.

Observe that when $A'\in T_A$ and $B'\in T_B$ we have $A'\times B'=(A'\times B)\cap (A\times B')\in T^*.$ And the set $$\{A'\times B': A'\in T_A\land B'\in T_B\}$$ is a base for $T^*.$

If $\beta_A$ is a base for $T_A$ and $\beta_B$ is a base for $T_B$ then $\{A'\times B': A'\in \beta_A \land B'\in \beta_B\}$ is also a base for $T^*$.

Examples: 1. $A=B=\mathbb R$ with the usual topology on $\mathbb R.$ A base for the product topology $T^*$ on $\mathbb R^2$ is $\{I\times J: I,J\in \beta_{\mathbb R}\}$ where $\beta_{\mathbb R}$ is the set of bounded open real intervals. The topology $T^*$ on $\mathbb R^2$ is equal to the topology generated by the base of "open disks": If $C\subset \mathbb R^2$ then $C\in T^*$ iff for every $(x,y)\in C$ there exists $r>0$ such that $(-r+x,r+x)\times (-r+y,r+y)\subset C$ iff there exists $r'>0$ such that $C\supset \{(x',y'): {r'}^2>(x'-x)^2+(y'-y)^2\}.$

1. Suppose $B=\{b\}.$ Then $A\times B,$ with the product topology, is homeomorphic to $A.$ The projection $p_A$ is not only continuous, it is a homeomophism.

2. Let $A=B=S^1$...... $S^1$ is standard notation for $\{(x,y)\in \mathbb R^2: x^2+y^2=1\},$ with the usual topology on $S^1$ as a subspace of $\mathbb R^2,$ where $\mathbb R^2$ has the product topology of Example 1..... With the product topology, $S^1\times S^1$ is homeomorphic to the surface of a donut.