Open Set Clarification: Sequence Definition

Question

Given a point, $x$, in a metric space $(X, d)$. I am trying to show the two bullets below are tied together by an iff.

• $\forall \epsilon>0, \exists N\in \mathbb{N}: (n\in \mathbb{N} \text { and } n\geq N\rightarrow d(x_n,x)<\epsilon$)
• $\forall$ open set $U$ containing $x$, $\exists N\in \mathbb{N}: [(n\in \mathbb{N} \text { and } n\geq N) \rightarrow x_n\in U$]

I worked out the problem this far, but I am confused as far by the terminology $\forall$ open set $U$ containing $x$, as I think it would refer to $\forall U\ni x$, which does not make much sense to me.

$\textbf{Question:}$ How do get $\bigcup_ {x\in B\subseteq U} B$ to just be $U$ with these quantifiers? I feel like I am misinterpreting something with my notation here and making this problem more difficult than it needs to be.

\begin{align*} \forall \epsilon>0, \exists N\in \mathbb{N}: (n\in \mathbb{N} \text { and } n\geq N\rightarrow d(x_n,x)<\epsilon)&\leftrightarrow \exists N\in \mathbb{N}: (n\in \mathbb{N} \text { and } n\geq N\rightarrow d(x_n,x)<\delta )\\ &\leftrightarrow \exists N\in \mathbb{N}: (n\in \mathbb{N} \text { and } n\geq N\rightarrow x_n\in B(x; \delta) )\\ &\leftrightarrow \exists N\in \mathbb{N}: (n\in \mathbb{N} \text { and } n\geq N\rightarrow x_n\in \bigcup_{x\in B\subseteq U} B ). \end{align*}

Answer 1

Notice that the union you wrote, $\bigcup_ {x_n\in B\subseteq U} B$, is $U$, because in particular $U \subseteq U$ and it contains some $x_n$ of your sequence. No "bigger" set can be in this union since this would imply that it is not contained in $U$. I believe this answers your question alone, though does not show the iff between the two bullets.

Answer 2

The first to the second follows from the fact that in a metric space by definition we have: $x \in U$ and $U$ open implies that there is some $\varepsilon >0$ with $B(x,\varepsilon) \subseteq U$. Then we apply the first bullet to this $\varepsilon$ and we're done, as $d(x_n,x) < \varepsilon$ iff $x_n \in B(x,\varepsilon)$ etc.

For the second to the first we only need to know that for every $\varepsilon>0$, the set $B(x,\varepsilon)$ is an open set containing $x$, so we can apply the second bullet to this set as our $U$.