# union(new: intersection) of any number of open sets is also open [closed]

I've just begun reading Spivak's Calculus on Manifold and attempted to proof this simple result.

-I've updated my proof-

My proof are as follows,

My proof for the intersection case still looks kinda dubious though.

@Devan Ware, the notation $N_{\epsilon}(x)$ looks very useful to me but i haven't seen it anywhere, which branch of math is it found in and where can i learn more about it?

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## closed as off-topic by arjafi♦Aug 17 '14 at 13:31

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Just let take $x\in\bigcup_{i\in I}G_i$ where $G_i$ open. Then $x\in G_i$ for some $i$ and hence $\exists N_{\epsilon}(x) \subset G_i \subset \bigcup_{i\in I}G_i$. A more interesting question to consider is that a finite intersection of open sets is also open. (just notice that if $x\in\bigcap_{i\in I}G_i$ ($I$ finite) then $\exists N_{\epsilon_i}(x)\subset G_i$ for each $i$ and so take $k = \min{\epsilon_i}$ then $N_{k}(x) \subset \bigcap_{i\in I}G_i$.) – Deven Ware Dec 3 '11 at 7:24
Ah, I'm sorry if the notation was a bit confusing but it means "Neighborhood of radius $\epsilon$" or "ball of radius $\epsilon$" it is from analysis (I learned it from baby Rudin) – Deven Ware Dec 3 '11 at 19:05
This question appears to be off-topic because it no longer has any content since the link to the image is dead. – arjafi Aug 17 '14 at 13:31

Your proof doesn't seem to be quite correct. Note that as your definition you have that $U \subseteq \mathbb{R}^n$ is open iff for each $x \in U$ there is an open rectangle $A = (a_1,b_1) \times \cdots \times (a_n,b_n)$ containing $x$ such that $A \subseteq U$. This means that for every point $x$ of $U$ you have to find such an open rectangle, and the choice of rectangle may depend on the choice of $x$. Thus, "picking $A$ to work for $U$ and $B$ to work for $V$" doesn't quite make sense. What you need to do is first pick the $x$ from the set you wish to show is open, and then show that there is an open rectangle that would work for this particular $x$.

The "trick" is to note that if $\{ U_i : i \in I \}$ is any family of sets, then $x \in \bigcup_{i \in I} U_i$ iff there is an $i \in I$ such that $x \in U_i$, and also note that if $A \subseteq U_i$ for some $i$, then $A \subseteq \bigcup_{i \in I} U_i$.

I think this should lead you in the right direction.

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For the infinite case, if the family is uncountable, then we cannot use mathematical induction. Here is the proof for the general case (whether it is finite or infinite, countable or uncountable): Let $\{U_i\}_{i\in I}$ be a family of open sets. Here $I$ can be finite, infinite, countable, or uncountable. We want to prove that $\cup_{i\in I}U_i$ is open.

To prove this, let $x\in\cup_{i\in I}U_i$. Therefore, $x\in U_i$ for some $i\in I$. Since $U_i$ is open by assumption, there exists an open rectangle $(a_1,b_1)\times(a_2,b_2)\times\cdots\times (a_n,b_n)$ such that $$x\in (a_1,b_1)\times(a_2,b_2)\times\cdots\times (a_n,b_n)\subset U_i.$$ This implies that $$x\in (a_1,b_1)\times(a_2,b_2)\times\cdots\times (a_n,b_n)\subset U_i\subset\cup_{i\in I}U_i.$$ Since $x$ is arbitrary, we have proved that $\cup_{i\in I}U_i$ is open.

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