Let \( U_{\alpha}\) \( (\alpha \in A) \) be a collection of open sets in \( {\mathbb R}^2.\) If \(A\) is finite, then the intersection \( U = \bigcap\limits_\alpha U_{\alpha} \) is also an open set. Here is a proof:

Suppose \(x\in U.\) For each \( \alpha \in A,\) let \( B_{\alpha}\) be a ball of some positive radius around \(x\) which is contained entirely inside \( U_{\alpha}.\) Then the intersection of the \( B_{\alpha}\) is a ball \(B\) around \(x\) which is contained entirely inside the intersection, so the intersection is open.

\((\)Here a *ball around \(x\)* is a set \( B(x,r)\) (\(r\) a positive real number) consisting of all points \( y\) such that \(|x-y|<r.\) In \( {\mathbb R}^2\) it is an open disk centered at \(x\) of radius \(r.)\)

Where does this proof go wrong when \(A\) is infinite?

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