# Product topology, subbase and weakness

Given two spaces $$(X_1,T_1), (X_2,T_2)$$ my book says that the product topology is the weakest topology s.t. the projection maps $$\pi_1:X_1 \times X_2 \rightarrow X_1$$ and $$\pi_2:X_1 \times X_2 \rightarrow X_2$$ are continuous. The base of the product topology $$T$$ on $$X_1 \times X_2$$ is defined as $$\{U_1 \times U_2: U_1\in T_1, U_2 \in T_2 \}$$.

Now I get the part where we take open sets $$U_1\in T_1$$ and show that $$\pi_1^{-1}(U_1)=U_1\times X_2$$ is open and similary for $$\pi_2^{-1}(U_2)=X_1 \times U_2$$. Then the intersection is $$U_1 \times U_2$$ and we can form all open sets in the product topology as the union of sets that are finite intersections of the form $$\pi_1^{-1}(U_1) \cap \pi_2^{-1}(U_2)$$. Then $$\{\pi_1^{-1} (U): U\in T_1 \} \cup \{\pi_2^{-1} (U): U\in T_2 \}$$ is a subbasis. I get all the points until here.

Now what I don't understand is the last part: "This makes $$T$$ the weakest topology for which the projection maps $$\pi_1$$ and $$\pi_2$$ are both continuous". We have not defined weakness this way, rather we have just said that $$T_1$$ is weaker than $$T_2$$ iff $$T_1\subset T_2$$ (on the same set $$X$$) that is that every open set in $$(X, T_1)$$ is also open in $$(X,T_2)$$. Just using this definition I cannot understand how we get the last sentence and would appreciate your help!

• You have to show that if $T$ is a topology which makes the projections continuous then the product topology is weaker than $T$.
– PtF
Commented May 15, 2021 at 9:59

## 2 Answers

Let $$\tau$$ be a topology on $$X_1\times X_2$$ wich is weaker than $$T$$. So, $$\tau\varsubsetneq T$$. Since $$T$$ is defined the way it is, this means that there is some open subset $$U_1$$ of $$X_1$$ such that $$U_1\times X_2\notin\tau$$ or that there is an open subset $$U_2$$ of $$X_2$$ such that $$X_1\times U_2\notin\tau$$. If the first possibility occurs, then $$\pi_1$$ is discontinuous, and if the second possibility occurs, then $$\pi_2$$ is discontinuous.

• Thank you for your clear and helpful reply. Commented May 15, 2021 at 13:45
• I'm glad I could help. Commented May 15, 2021 at 13:48

If $$\mathcal{T}$$ is any topology on $$X_1 \times X_2$$ so that both

$$\pi_1: (X_1 \times X_2, \mathcal{T}) \to (X_1, \mathcal{T}_1) \text{ and } \pi_2: (X_1 \times X_2, \mathcal{T}) \to (X_2, \mathcal{T}_2)$$

are continuous, then the continuity of $$\pi_1$$ forces that $$\pi_1^{-1}[U] \in \mathcal{T}$$ for all $$U \in \mathcal{T}_1$$ and also that $$\pi_2^{-1}[U] \in \mathcal{T}$$ for all $$U \in \mathcal{T}_2$$ follows from the other continuity. So the subbase you mentioned is a subset of $$\mathcal{T}$$ and so the base the subbase generates too: $$\{U_1 \times U_2\mid U_1 \in \mathcal{T}_1, U_2 \in \mathcal{T}_2\} \subseteq \mathcal{T}$$ and it follows that $$\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}$$

So for any topology that makes both $$\pi_i$$ continuous, $$\mathcal{T}_{\text{prod}}$$ is a subset of it. This is exactly the minimality that your text is talking about. (weaker = subset). We have just enough open sets in it to make the projections continuous and nothing more. This minimality gives it a lot of nice properties. It also shows how to generalise to infinite products.

• Thanks for your answer! Helped a lot. Commented May 15, 2021 at 13:45