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so my professor mentioned that when a normed space is finite dimensional the norm and weak topologies agree. To show the topologies agree it should be enough to show that they have the same convergent nets. I can see how if a net $x_\alpha\rightarrow x$ in the norm topology then it converges in the weak topology. Namely, for arbitrary $f\in X^*$ we know by continuity that $f(x_\alpha)\rightarrow f(x)$, which means $x_\alpha\rightharpoonup x$. However, I can't see the other way. Would anyone have some intuition as to why convergence in the weak topology implies convergence in the norm topology? I'm assuming that this direction is where the finite dimension of our space comes in to play. Thanks in advanced.

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We can identify our space $X$ with $\mathbb{R}^n$, with the norm topology being the usual Euclidean topology on $\mathbb{R}^n$ (since all norms on a finite-dimensional space are equivalent). Now if $x_\alpha\to x$ weakly, then $f_i(x_\alpha)\to f_i(x)$ where $f_i:\mathbb{R}^n\to\mathbb{R}$ is the $i$th coordinate functional. But that just means that $x_\alpha$ converges to $x$ on each coordinate, which implies $x_\alpha$ converges to $x$ in the Euclidean topology. Thus $x_\alpha\to x$ in the norm topology.

More generally, a finite-dimensional vector space only admits one topological vector space structure (i.e., only one Hausdorff topology such that addition and scalar multiplication are continuous). This is a bit more complicated to prove; see this answer for instance

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  • $\begingroup$ So this proof fails in the infinite dimensional case as we can't identify our space with $\mathbb{R}^n$, correct? $\endgroup$ – Scott Jan 27 at 23:16
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Eric Wofsey Jan 27 at 23:19
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Another way to do this is to just go back to the definitions. Let $\epsilon>0,$ and $x_0\in X$. It suffices to show that any open ball $B(x_0,\epsilon)$ is weakly open. We may use $\|\cdot\|_{\infty}$ as our norm. And since translation is a homeomorphism, we may take $x_0=0$. Choose a basis $(e_i)^n_{i=1}$ for $X$. Then, $x=\sum^n_{i=1} x_ie_i$ for each $x\in X$ and for each $1\le i\le n,$ we have the functionals $e_i^*:X\to \mathbb C:x\mapsto x_i.$

Then, $B_{\epsilon }=\left \{ x:\left \| x\right \|<\epsilon \right \}=\left \{ x:\forall\ 1\le i\le n,\ |x_i|<\epsilon \right \},\ $ which is nothing more than the weakly open set $\left \{ x:\forall\ 1\le i\le n,\ |e_i^*(x)|<\epsilon \right \}.$

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