Often weak convergence is defined as follows:
Def 1: A net $(x_\alpha)\subseteq X$ converges weakly to $x\in X$ if $\langle x_\alpha, x^* \rangle \rightarrow \langle x, x^*\rangle$ for all $x^* \in X^*$
Conway's A Course in Functional Analysis however first defines the weak topology as the topology on $X$ generated by subbase-elements of the form $\{x\in X: |\langle x, x^* \rangle| < \varepsilon\}$. Then weak convergence of a net $(x_\alpha)\subseteq X$ is defined as convergence of the net in the weak topology, which can be rigorously stated as
Def 2: A net $(x_\alpha)\subseteq X$ converges weakly to $x\in X$ if for every weakly open neighborhood $U\subseteq X$ of $x$ there exists $\alpha_0$ such that $x_\alpha\in U$ for all $\alpha \geq \alpha_0$.
My question is: why are these definitions equivalent?
I first tried to show that definition 2 implies definition 1, for a net $(x_\alpha)$ converging to $0$. For the weakly open $U$ there exist $\varepsilon_1, …, \varepsilon_n > 0$, and $x_1^*, …, x_n^* \in X^*$ such that
$$V:= \bigcap_{i=1}^n\{x\in X: |\langle x, x_i^*\rangle| < \varepsilon_i\} \subseteq U.$$
Then intuitively speaking, shrinking $U$ would necessitate shrinking $V$, thus making the $\varepsilon_i$ smaller and thus letting the $|\langle x, x_i^*\rangle|$ go to $0$. I don't understand however how we can relate this to convergence of the net $(\langle x_\alpha, x^*\rangle) \subseteq \mathbb{F}$ (where $\mathbb{F}$ denotes the relevant field) to zero for every $x^*\in X^*$.
So just to be clear, this question can be regarded as being twofold:
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How does convergence of the net $(\langle x_\alpha, x^*\rangle) \subseteq \mathbb{F}$ follow from the $x_\alpha$ being contained in $U$ for all $\alpha \geq \alpha_0$
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Why does this convergence of $(\langle x_\alpha, x^*\rangle)$ hold for all $x^*\in X^*$? (probably the answer to this question is contained in the answer to the first question.)
Maybe I'm overseeing some things that are obvious; working both with nets and the weak topology is quite new to me, so I'm probably just getting too distracted by technical details.
Best Answer
A net $(t_\alpha)$ in $\mathbb R$ converges to $t$ if and only if for all neighborhoods $U$ of $t$ there is $\alpha_0$ such that $t_\alpha \in U$ for all $\alpha \ge \alpha_0$.
Since we are in topological vector spaces, it suffices to prove the claim for nets $(x_\alpha)$ converging weakly to $0$.
Let $(x_\alpha)$ converge according to definition 1 to $0$. Let $U$ be a neighborhood of $0$, thus it contains a subset $V$ given by $$ V := \{ x : \ |x_i^*(x)|\le \epsilon_i\}. $$ Now according to definition 1, for all $i$ there is $\alpha_i$ such that $|x_i^*(x_\alpha)|\le \epsilon_i$ for all $\alpha \ge \alpha_i$. Take $\alpha_0$ such that $\alpha_0\ge \alpha_i$ for all $i$. Then $|x_i^*(x_\alpha)|\le \epsilon_i$ for all $\alpha \ge \alpha_0$ and all $i$. Hence, $x_\alpha\in V \subset U$ for all $\alpha \ge \alpha_0$. And we have weak convergence according to definition 2.
Now let $(x_\alpha)$ converge according to definition 1 to $0$. Take $x^*\in X^*$, $\epsilon>0$. Then there is $\alpha_0$ such that $$ x_\alpha \in \{ x: \ |x^*(x)|\le \epsilon\} \quad \forall \alpha\ge\alpha_0. $$ So $|x^*(x_\alpha)|\le \epsilon$ for all $\alpha\ge\alpha_0$, which is convergence according to definition 1.