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Prove that the union of any (even infinite) number of open sets is open.

My attempt:

Let $A_1,...,A_n$ be open sets such that $n=\{1,...,\infty\}$.

Let $U=\cup_{i=1}^\infty A_i$.

Take any $x\in U$. By definition of union, $x\in A_1,...A_n$.

$A_1,...A_n$ is open so there exists some open ball $Br(x)\subset U$ that contains $x$. Thus $U$ is open.

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    You're confusing *union* and *intersection*. It is false that the intersection of an infinite number of open sets must be open. – 2017-01-28
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    What is your definition of "open"? – 2017-01-28
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    @TedShifrin got it, so $x\in A_1,...,A_n$ is false. – 2017-01-28
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    $x \in \bigcup\limits_{i=1}^{\infty} A_i$ means that $x$ is in at least one of the $A_i$. – 2017-01-28
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    so can I just say that since all the $A_i's$ are open, they contain an open ball. So their union will also contain an open ball. Thus, all $x\in U$ will be contained in this open ball and therefore $U$ is open. – 2017-01-28
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    "In this open ball"? I hope you don't mean one and the same open ball... – 2017-01-29
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    Are you working in metric space? – 2017-01-29

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You should use an arbitary index set if you want an arbitrary union: suppose $A_i, i \in I$ is a family of open sets (in a metric space, I see from your proof attempt). Using the definition

$$O \text{ is open iff } \forall x \in O: \exists r>0 : B_r(x) \subseteq O\text{,}$$

as you seem to be doing.

To see $A = \bigcup_i A_i$ is open, we take any $x \in A$. By the definition of union, there exists some $i_0 \in I$: $x \in A_{i_0}$.

As $A_{i_0}$ is open and $x \in A_{i_0}$ we have some $r >0$ such that

$$B_r(x) \subseteq A_{i_0} \subseteq \bigcup_{i \in I} A_i = A$$

where the last statement is a set theory triviality (any set in a union is a subset of the union). So we have shown indeed that for any $x \in A$ we have this ball $B_r(x)$ with $B_r(x) \subseteq A$ (we just re-use the one we get for free from $A_{i_0}$ being open). So $A$ is open by definition.