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The proof is here: (link).

I don't see how the third line (starting with Thus: $\exists \epsilon_i$...) is justified. That is: just because $x \in U_i$, for all $i$, how do I know that a neighborhood of $x$ is in $U_i$, for all $i$?

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In the context of a metric space, a subset $U$ is open when for any $x$ in $U$, one can find an $\epsilon$-ball centered at $x$ and contained in $U$.

The third line applies this definition to find a collection of open balls centered at $x$ and contained in the respective $U_i$.

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This is just by definition: the $U_i$ are open, so there is an $\epsilon_i$-ball around $x$ that is contained in $U_i$.

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  • $\begingroup$ Okay: I understand it now. There is some neighborhood in each $U_i$ for that x, thus we just pick the smallest one, and then it HAS to be in all the others. Thanks. $\endgroup$ – user156349 Jun 10 '14 at 17:14
  • $\begingroup$ @user156349 Yup! You got it. $\endgroup$ – user98602 Jun 10 '14 at 17:15

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