# Don't understand proof that union of open sets is open

All of the proofs I read of this are pretty much the same, but they all don't explain the last part of the proof and end it there because I assume they think it is obvious to everyone.

The proofs usually go like: Let $S = \bigcup_{i \in I} S_i$ for open sets $S_i$. Prove that $S$ is open. If $x$ is in some $S_i$, since $S_i$ is open, $\forall x \in S_i$ $\exists r > 0$ such that $N_r(x) \subset S_i \subset \bigcup\limits_{i \in I} S_i = S$. Thus $S$ is open.

I don't understand the final implication. We want to show that in $S$, every element of all sets are possible to be the center of an open ball around them with some radius that is contained within $S$. We've only stated that some $S_i$ is an open subset of $S$, which is actually a given so we haven't proven anything. Can someone explain the final part?

• Yes but $S_i$ is contained in $S$ so the open ball is also in $S$ Feb 22, 2015 at 1:22

The point is that the open ball $N_r(x)$ is not only a subset of some $S_i$, it is also a subset of $S$. The transitive property of the subset relation is being used.
• I understand that the open ball $N_r(x)$ is also a subset of the larger $S$. But I fail to see why this makes $S$ open, because in order for $S$ to be open, every element of $S$ needs to be the center of an open ball contained in $S$, and so far all we've done is only shown that there exists one ball in $S$ so far (the $N_r(x)$ one). Feb 22, 2015 at 1:27
• @user130018: What do you want more? The existence of an open ball around every $x$ that is contained in $S$ is exactly the defintion of $S$ being open. You're even quoting that definition yourself in the same breath. You seem to be saying "we need to prove such-and-such but so far all we've done is to prove such-and-such". Feb 22, 2015 at 1:29
• Is your issue that we only seem to have found a ball in $S$ for only one point? Feb 22, 2015 at 1:31
• @user130018 I think your confusion stems from not recognizing that we started by picking an arbitrary $x\in S$ and proceed to show that $N_r(x)\subseteq S$. This means that every $x\in S$ satisfies $N_r(x)\subseteq S$, and it is this that implies that $S$ is open - i.e. we have discharged the $x$ in the last line of the proof. Feb 22, 2015 at 1:32
• @MarioCarneiro: Well, we're not really "picking" $x$ -- the adversary picks one for us, and we're just reacting to his pick. Feb 22, 2015 at 1:34