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Technically a metric space is not a topological space, and a topological space is not a metric space: a metric space is an ordered pair $\langle X,d\rangle$ such that $d$ is a metric on $X$, and a topological space is an ordered pair $\langle X,\tau\rangle$ such that $\tau$ is a topology on $X$. A metric on $X$ is a special kind of function from $X\times X$ ...

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Let $||\cdot||_0$ and $||\cdot||_1$ induce the same topology on a vector space $V$. Let $r>0$, since $B_r(0)_0$ is open in the topology induced by $||\cdot||_0$, it must be open in the one induced by $||\cdot||_1$. This means there exists a $k>0$ so that $B_k(0)_1 \subset B_r(0)_0$. Now any $x \in V, x \neq 0$ can be written as ...

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The weak-* topology on $H^*$ is generated by the functionals $$\{h^* \mapsto h^*(x) \mid x \in H\}$$ If you pull-back this topology to $H$, you get the topology generated by $$\{y \mapsto \Phi(y)(x) \mid x \in H\} = \{y \mapsto (x,y) \mid x \in H\}$$ But this is the same as the weak topology which is generated by $$\{y \mapsto (y,x) \mid x \in H\}$$

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This question and answer give a concrete example of a sequence $(\delta_n)$ in a compact space that has no convergent subsequence (so this compact space is not sequentially compact). The answer gives a "concrete" (if you believe ultrafilters are concrete) subnet $(x_d), d \in D$ that converges to some $f_\mathcal{U}$. This subnet cannot have a subsequence ...

2

$\mathscr D_K \cap W = \bigcap_{x_{j}\in K} \lbrace\varphi \in \mathscr D_K: |\varphi(x_j)| <c_j\rbrace$ for some finitely many $x_{m}$ in $K$ and corresponding constants $c_j>0$ (given explicitly in Rudin's proof). So all you need is the continuity of the evaluation $\mathscr D_K\to \mathbb C$, $\varphi\mapsto \varphi(x)$. By the way, I do not like ...

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By definition, unitaries preserve the norm. So $$\|Cy-CSy\|=\|C(y-Sy)\|=\|y-Sy\|.$$ Thus, $CV=V$.

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Suppose that $(X,\mathcal{U})$ is sequentially compact. So there exists some $N \in \mathcal{U}$ (some entourage) such that $X$ cannot be covered by finitely many $B(x,N)$. So pick $x_0 \in X$, note that $B(x_0, N)$ does not cover $X$ so there is some $x_1 \notin B(x_0,N)$. By recursion, if we have constructed points $x_0,\ldots, x_n$ then we find some ...

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The map $\mathbf{R}\times X\to X$, $(r,x)\mapsto x$, is continuous by assumption, with $\mathbf{R}$ given the usual topology. As a consequence, the map $\mathbf{R}\to X$, $r\mapsto rx_0$, is continuous. Since the sequence $1/n$ converges to $0$, you have $$\lim_{n\to\infty}\frac{1}{n}x_0=0x_0=0$$ Well, if $X$ is not Hausdorff, $\lim$ is used in the sense ...

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Every metric space is a topological space. A subset $S$ of a metric space is open if for every $x\in S$ there exists $\varepsilon>0$ such that the open ball of radius $\varepsilon$ about $x$ is a subset of $S$. One can show that this class of set is closed under finite intersections and under all unions, and the empty set and the whole space are open. ...

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A topological vector space in which there is a bounded neighbourhood of $0$ is called locally bounded. Whether the TVS $X$ is $T_1$ or not doesn't play much of a role, $X$ is locally bounded if and only if its associated Hausdorff TVS $X/\overline{\{0\}}$ is locally bounded. If we're only considering vector spaces over $\mathbb{R}$ or $\mathbb{C}$, it ...

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