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Let $(X, \tau)$ be a topological space. Given a subset $A$ and a collection of open sets $U_i$, $i \in I$ for some index set, s.t. $A \subseteq U_i$ for all $i \in I$. How can I build smallest open set that contains $A$?

Intersection of infinite number of open sets in not necessarily open, so we can't assume that $U = \cap_{i \in I}U_i$ is open. But $U = Int(\cap_{i \in I}\overline{U_i})$. Intersection of closed sets is closed and interior is open. I think that interior is not empty in this case, but can't prove it. Since $A \subseteq U_i$, then $A \subseteq U$ ?

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    $\begingroup$ It's worth trying an example for this. What happens if you try to find the smallest open set in $\mathbb R$ containing $A=\{0\}$? $\endgroup$ – Milo Brandt Aug 19 '18 at 23:43
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Consider the real line with the usual topology, $A=[0,1]$ and $U_i=(-1/i,1+1/i), i\in \mathbb{N}.$

Firt of all note that there is no a smallest open set containing $A.$ Indeed $A\subset (-\epsilon, 1+\epsilon), \forall \epsilon >0.$

Now, $U=Int(\cap_{i \in I}\overline{U_i})=(0,1)$ which doesn't contain $A$.

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