Weak topology on $\mathbb{R}$

Question

It is known that $\mathbb{R}$ is a Hilbert space.

My question is: does weak and strong topology are the same thing in $\mathbb{R}$? If not, what does mean weak topology on $\mathbb{R}$?

Thank you in advance!

Answer 1

Let $X$ be a Banach space (this includes $\mathbb{R}$, Hilbert space). Consider the space of all continuous linear functional on $X$ and denote it by $X^*$.

You already have a topology on $X$ which is induced by its metric linear (here the metric means norm) structure. However, sometimes this topology is too strong (contains more open sets) that are required to maintain the continuity of each members of $X^*$. The weakest topology (the least open sets) that is needed to maintain the continuity of each members of $X^*$ is called weak topology on $X$.

However, in case of finite-dimensional Banach spaces, the norm topology and the weak topology coincides. Thus, in your case $\mathbb{R}$ with weak topology is nothing but the usual topology on $\mathbb{R}$ ($\mathbb{R}$ is a one-dimensional space over itself), induced by the modulus metric.

Answer 2

All functionals on $\Bbb R$ are of the form $f(x)=ax$ for some $a \in \Bbb R$. The minimal topology $\mathcal{T}_w$ on $\Bbb R$ that makes all of these continuous (when the codomain has the usual topology $\mathcal{T}_e$) is of course $\mathcal{T}_e$: it's clear that $\mathcal{T}_w \subseteq \mathcal{T}_e$ as the latter is one topology that makes all these functionals continuous. And as $1_{\Bbb R}:=f(x)=x$ is just one of these functionals too, $1_{\Bbb R}: (\Bbb R,\mathcal{T}_w) \to (\Bbb R,\mathcal{T}_e)$ must be continuous so that $\mathcal{T}_e \subseteq \mathcal{T}_w$ by definition of continuity...

So two inclusions yield equality: $\mathcal{T}_e= \mathcal{T}_w$.