1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ You have to show that if $T$ is a topology which makes the projections continuous then the product topology is weaker than $T$. $\endgroup$
    – PtF
    Commented May 15, 2021 at 9:59

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you for your clear and helpful reply. $\endgroup$ Commented May 15, 2021 at 13:45
  • $\begingroup$ I'm glad I could help. $\endgroup$ Commented May 15, 2021 at 13:48
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer! Helped a lot. $\endgroup$ Commented May 15, 2021 at 13:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .