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In other to show that

The intersection of a finite number of open sets is open.

We are doing the following:

Let $Q_1, ..., Q_n$ be a open sets, and $x \in \bigcap_{i=1}^n Q_i$ be arbitrary. So by definition $\exists r_i>0$ s.t $B(x,r_i) \subset Q_i$, so define $r = min\{r1,...,r_n\}$, hence $$B(x,r) \subset B(x, r_i) \subset Q_i \quad \forall i$$, hence $$B(x,r) \subset \in \bigcap_{i=1}^n Q_i,$$ and the intersection is open.

So my question is that if it was not a finite but an infinite intersection of subset, which step of the above proof would fail ?

Our
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  • $r$ would be a mininum over an infinite set, it could be zero. For example if $r_i=1/i$. – vap Oct 19 '17 at 06:16
  • The minimum over an infinite set does not need to exist, for example ${1/n \mid n \in \mathbb{N}}$. – Joppy Oct 19 '17 at 06:20

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Consider real numbers and $Q_i=(-1/i;1/i)$. But $$ \bigcap_i Q_i=\{0\} $$ and there is no space for a ball.

Przemysław Scherwentke
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  • Possible answers:
    https://math.stackexchange.com/questions/339898/why-can-the-intersection-of-infinite-open-sets-be-closed and https://math.stackexchange.com/questions/1460853/infinite-intersection-of-open-sets     – Zbigniew Oct 19 '17 at 06:27