Consider a set $X$ with topologies $\tau_1$ and $\tau_2$. Let $\tau=\tau_1\lor\tau_2$ be the collection of subsets $W\subseteq X$ such that for every $x \in W$ there are subsets $U,V \subset X$ with $x \in U$, $x\in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$. This question has three parts.

1.) Show that $\tau$ is a topology on $X$.

For this part, I know the properties that I have to show. It is clear that both $\emptyset \in \tau$ and $X \in \tau$. I know that I need to show that finite intersections and arbitrary unions remain in $\tau$. To start, I let $W_1,W_2, . . ., W_n \in \tau$. If $W_i = \emptyset$ for any $1 \leq i \leq n$, then the intersection is $\emptyset$, and the union is unaffected, so we can assume that each $W_i$ is nonempty. I don't know where to go from here.

2.) Show that both $\tau_1 \subseteq \tau$ and $\tau_2 \subseteq \tau$.

This seems to me to be trivial, so I am afraid that I might be missing something here. Some guidance in the right direction other than a direct answer would be appreciated.

3.) If $\tau'$ is a topology on $X$ such that $\tau_1,\tau_2 \subseteq \tau'$, then $\tau \subseteq \tau'$.

Let $W \subseteq \tau$. Then, for every $x \in W$ there are subsets $U,V \subset X$ with $x \in U$, $x\in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$. But then since $\tau_1,\tau_2 \subseteq \tau'$, we have $U,V \in \tau'$, and thus $U \cap V \subseteq \tau'$. I'm not sure how to jump from here to saying that $W \subseteq \tau'$.


First of all, we can make a small adjustment which might make the problem clearer. An equivalent definition of $\tau$ is as follows. Given $W\subseteq X$, we have $W\in\tau$ if and only if $$W=\bigcup_{x\in W}(U_{(W,x)}\cap V_{(W,x)}),$$ for some $U_{(W,x)}\in\tau_{1}$ and $V_{(W,x)}\in\tau_{2}$ which satisfy $x\in U_{(W,x)}$ and $x\in V_{(W,x)}$ (for each $x\in W$). This may seem like an odd way of proceeding but it makes the questions a bit easier.

With this new formulation, it should be clear that an arbitrary union of elements of $\tau$ is also in $\tau$ (given a collection of $W_{i}$'s in $\tau$, each can be written as a union of $U\cap V$'s, take the union of all of these unions and what do you get?). For the intersection property, given $U,U'\in\tau_{1}$ and $V,V'\in\tau_{2}$ with $U\cap V\subseteq W$ and $U'\cap V'\subseteq W'$ (where $W,W'\in\tau$), note that $$U\cap V \cap U' \cap V'=(U\cap U')\cap(V\cap V')\subseteq W\cap W'.$$

This should help you prove the intersection axiom.

You are right that 2) is very easy. In the above reformulation, I claim that for each $W\in\tau_{1}$, we can just write $W=U\cap V$ for some $U\in\tau_{1}$ and some $V\in\tau_{2}$ (you probably came up with a similar idea). This shows that $\tau_{1}\subseteq\tau$. A similar method shows $\tau_{2}\subseteq\tau$.

Finally for 3), the final jump you need to make comes from the new formulation I gave in the first paragraph. You have shown that for each $x\in W$ we can find $U,V\in\tau'$ with $x\in U\cap V$ and $U\cap V\in\tau'$. Now let $C_{(W,x)}=U\cap V$ for these $U,V$ (which depended on $W$ and $x$), and you have shown that $C_{(W,x)}\in\tau'$. Taking the union over all $x\in W$ of the $C_{(W,x)}$ gives $W$, and shows that $W\in\tau'$ (can you see why?).

EDIT: I will just add that there is another way of viewing this topology $\tau_{1}\vee\tau_{2}$. If you know anything about topological bases, then you can observe that $\tau_{1}\vee\tau_{2}$ is the topology generated by the topological base $$\mathcal{B}=\{U\cap V:U\in\tau_{1},V\in\tau_{2} \}.$$ It is easy to check that $\mathcal{B}$ is indeed a base for a topology, and the topology it generates is precisely $\tau_{1}\vee\tau_{2}$.

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  • $\begingroup$ Thank you for your very detailed answer! Your reformulation is very helpful! I think I have 1) down now using the new formulation, and I had this already for 2). For 3), since $\tau'$ is a topology, the union of elements of $\tau'$ must also be an element of $\tau'$, and so $W \in \tau'$, correct? $\endgroup$ – mathqueen459 Feb 19 '18 at 22:21
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    $\begingroup$ @mathqueen459 Yes that's correct. A common trick in topology is, if you have a set $W$ and you can show for each $x\in W$ there is an open set $O_{x}$ such that $x\in O_{x}$ and $O_{x}\subseteq W$, then you can write $W=\cup_{x\in W}O_{x}$. Thus, $W$ is open as it is a union of open sets. $\endgroup$ – JonCC Feb 20 '18 at 15:52

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