As in the title, if $X$ is infinite dimensional, all open sets in the $\sigma(X,X^{\ast})$ topology are unbounded. The $\sigma(X,X^{\ast})$ topology is the weakest topology that makes linear functionals on $X^\ast$ continuous. How does one show this? How does having an infinite basis relate to open sets being unbounded? I can't see this, please help and thanks in advance!
If $X$ is infinite dimensional, all open sets in the $\sigma(X,X^{\ast})$ topology are unbounded.
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$\begingroup$
general-topology
functional-analysis
vector-spaces
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7Hint: The basic open neighborhoods of $0$ are of the form $\bigcap_{i=1}^n \{x : \lvert \varphi_i(x)\rvert \lt \varepsilon_i\}$, so such a neighborhood contains the subspace $\bigcap_{i=1}^n \ker{\varphi_i}$. – 2012-10-29
1 Answers
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It's enough to show it for basic non-empty open sets which contain $0$ (for the others, do a translation). These ones are of the form $V_{N,\delta,f_1,\dots,f_N}=\bigcap_{j=1}^N\{x\in X, |f_j(x)|<\delta\},$ where $N$ is an integer, $f_j\in X^*$ and $\delta>0$, $1\leq j\leq N$. Then $\bigcap_{j=1}^N\ker f_j\subset V_{N,\delta,f_1,\dots,f_N}.$ As $X$ is infinite dimensional, $\bigcap_{j=1}^N\ker f_j$ is not reduced to $0$ (otherwise the map $x\in X\mapsto (f_1(x),\dots,f_N(x))\in\Bbb R^n$ would be injective). So it contains a non-zero vector $x_0$, and $\lambda x_0$ for all scalar $\lambda$, proving that $V_{N,\delta,f_1,\dots,f_N}$ is not bounded.
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0@MattN. :-)${}{}$ – 2012-10-29