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You are close, but are confusing the definition of your pseudometics: the elements of your spaces are equivalence classes of functions given by $\{y\in C^n:\tilde{d^j}(x,y)=0\}$. Again, $x$ and $y$ are functions in the preceding set. If it helps you remember/keep track, we could rewrite the classes as $\{g\in C^n:\tilde{d^j}(f,g)=0\}$, using variable letters ...


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No, it is not necessary that $\lVert\,\cdot\,\rVert$ belongs to the family $P$. Unless $X = \{0\}$, when there is only one seminorm on the space - we could work around that if we allow $P = \varnothing$. And if $\dim X = 1$, then all seminorms on $X$ are constant multiples of the norm, and we must put at least one non-zero multiple of the norm in the family. ...


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if $\{U_n\}$ is a countable neighborhood basis at $0$, is $\{U_n+U_n\}$ again a neighborhood basis at $0$? Yes. Let $U$ be an arbitrary neighborhood of the zero. There exists a neighborhood $V$ of the zero such that $V+V\subset U$, and there exists a base neighborhood $U_n$ of the zero such that $U_n\subset V$. This should allow you to apply ...


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One side is rather trivial, since the closure operator preserves inclusions: $$B \subset \overline{B} \implies A \cap B \subset A \cap \overline{B} \implies \overline{A \cap B}\subset \overline{A \cap \overline{B}}.$$ Now, for the other side. We have to use that $A$ is open. We do it like this: take $x \in \overline{A \cap \overline{B}}$ and $U$ an open ...



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