Showing that a set is open in a topological space

I have come across this exam question on a past paper.

$(X,T)$ is a topological space.
Suppose $A \subseteq X$ is such that for every $x \in A$ there is an open set $B$ such that $x\in B \subseteq A$. Show that $A$ is open.

Is it right to say that $A$ is open because it is the infinite union of sets $B$ which are open?
Is my reasoning enough to 'show' that $A$ is open ?

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Yes, your reasoning is enough. Denote $B_{x}$ as the open neighbourhood $B\subset A$ of $x$. What you need to show is that $A=\cup_{x\in A}B_{x}$. –  Thomas E. Apr 23 '12 at 16:20
The word "infinite" is unnecessary, though. –  Arturo Magidin Apr 23 '12 at 16:20

Yes, your idea is correct. For each $x \in A$, let $B_x \subseteq A$ be the corresponding open neighborhood of $x$. Then $$A = \bigcup_{x \in A} B_x.$$ Note that equality holds since $B_x \subseteq A$ for each $x$, and each point $x \in A$ is covered by the union. Since an arbitrary union of open sets is open, $A$ itself is open.

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The word "infinite" is unnecessary.

But it is true that $A$ is a union (possibly infinite, but that is irrelevant) of open sets, hence is open.

I'll try to avoid what seems like a use of AC (for each $x$, pick a $B_x$ with the appropriate properties). For each $x\in A$, let $$\mathcal{B}_x = \{B\subseteq X\mid B\text{ is open, }x\in B,\text{ and }B\subseteq A\}.$$ We know, by assumption, that $x\in\bigcup\mathcal{B}_x$ (since the set is not empty). (Recall that if $X$ is a set whose elements are sets, then $\bigcup X$ is the union of the elements of $X$). Moreover, $\bigcup\mathcal{B}_x$ is a union of open sets, hence open, and every element of $\mathcal{B}_x$ is contained in $A$, hence $\bigcup\mathcal{B}_x$ is contained in $A$.

Now consider $$B = \bigcup_{x\in A}\left(\bigcup\mathcal{B}_x\right).$$ This is open, being a union of open sets; contained in $A$, being a union of sets contained in $A$; and contains $A$, since for each $x\in A$, $x\in\bigcup\mathcal{B}_x\subseteq B$. Thus, $A=B$, so $A$ is open.

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+1: Good illustration of the standard trick of "don't pick one, pick all!" –  Zhen Lin Apr 23 '12 at 17:37
Fix $x \in A$. Since $x \in B$ and $B$ is open, there is a neighborhood $U$ of $x$ such that $x \in U \subset B$. But then $x \in U \subset A$. Therefore each point of $A$ is an interior point, and $A$ is open in $X$.