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I'm going over the proof of the theorem stating that "In a metric space, compactness impliess sequential compactness". I'm very likely confusing myself. I have the following proposition:

$\forall x\in X$, $\exists \epsilon(x)$ such that $B(x;\epsilon(x))$ contains only finitely many terms from the sequence $\{y_j\}$.

Now I have already proved that this is not the case (since this is true then the covering I found would actually miss infinitely many terms from the sequence). Then I tried to negate this proposition. Then I confused myself. Should it be like:

$\exists x\in X $ such that $!\exists \epsilon$ satisfying that $B(x;\epsilon)$ contains only finitely many terms.

But if this is true, what about a ball contains no term from the sequence?

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Two things.

  1. "Not exists $\varepsilon$ such that $B(x;\varepsilon)$ contains only finitely many terms" is the same as saying "For every $\varepsilon$, $B(x;\varepsilon)$ contains infinitely many terms". Which is just clearer.

  2. If it contains no sets, it contains $0$ terms, it most certainly contains only finitely many of them.

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I tend to write it as:

$\exists x\in X$ such that $\forall \epsilon$, $B(x;\epsilon)$ contains infinitely many terms from $\{y_j\}$.

Change all the $\exists$ to $\forall$ and vice versa.

If you're saying you've found a ball around $x$ that contains no terms from the sequence $\{y_j\}$, then you found a ball around $x$ that contains only finitely many terms from that sequence.

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