Weak topology net: $x_\alpha \rightarrow x $ iff $f(x_\alpha) \rightarrow f(x)$. Are open sets translation invariant?

104 Views Asked by At

As described in wikipedia,

If $\langle x_\alpha \rangle $ is a net in vector space $X$ with the weak topology, then $x_\alpha \rightarrow x$ iff $\phi(x_\alpha) \rightarrow \phi(x)$ for all $\phi \in X^*$.

I am stuck on the converse. I found this proof which I understood except the first line,

we see that $x_\lambda$ converges weakly to $x$ iff $x_\lambda-x$ converges weakly to $0$.


My thoughts: I think we need to show that open sets in $X$ are translation invariant - but I don't see how that is true in the weak topology.