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1

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 ...

1

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. ...

1

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 ...

0

Let $f$ be a non zero linear functional from the topological vector space $X$ to $\mathbb{C}$. It is enough to show that $f$ maps a balanced neighborhood of $0$ to an open ball around $0\in \mathbb{C}$ or the whole space $\mathbb{C}$. The former case corresponds to a continuous map $f$, the later to an unbounded map. Let $V$ be a balanced neighborhood ...

0

The vectot space $\mathcal C^\infty\bigl([0,1],\mathbf R\bigr)$ seems to be an example: you define its topology with the family of semi-norms: $$\lVert f\rVert_k=\sup_{x\in[0,1]}\bigl\lvert f^k(x)\bigr\rvert$$ It can be shown it is a Fréchet space for the topology defined by this family (hence it is metrisable), but its topology can't be defined by a single ...

1

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 ...

0

I assume you've proved $\|\,\|$ is a norm on $V.$ Suppose $f_n$ is Cauchy in this norm. Fix $x$. Then $$|f_n(x) - f_m(x)| = |f_n(x) - f_m(x)- (f_n(a) - f_m(a))| \le \text {Var} (f_n-f_m).$$ Hence $f_n(x)$ is Cauchy in $\mathbb {R}$ and thus converges. Thus $f_n \to$ some $f$ on $[a,b]$ pointwise everywhere. Now $(f_n)$ Cauchy in this norm implies $\sup_n ... 0 If by "second category" you mean "nonmeager," why would this be true? Take$A=\emptyset$:P, or less trivially$A=\{0\}$. Or, say, let$B=\mathbb{R}^2$,$A=\mathbb{R}\times\{0\}$with the usual topology - then$B$is nonmeager, but$A$is nowhere dense, yet clearly convex. 0 Here is a relatively concrete normed space:$l^\infty / c_0 =: E$. There are concrete nonzero elements, say the coset$(1,1,1,\cdots)+c_0$. But using only ZF, one cannot prove the existence of a nonzero bounded linear functional on$E$. 1 Remark$ 3 $on Page$ 8 $of this document says that it is an utterly doomed enterprise to establish the statement of the OP without using the Hahn-Banach Theorem ($ \mathsf{HB} $). If$ \mathsf{ZF} $is consistent, then$ \mathsf{HB} $is independent of$ \mathsf{ZF} $. Assuming that$ \mathsf{ZF} $is consistent, there exists a model of$ \mathsf{ZF} + ...

5

Let $p,q$ such as $1\leq p<q$ and $x = \{x_k\} \in l^p$. Then by definition $\sum_{k=1}^{\infty} {|x_k|^p} < \infty$. Therefore $|x_k| \to 0$ as $p \to \infty$, hence $|x_k| \to 0$. This implies that there exists an index $k_0$ such that $|x_k| < 1$ for any $k ≥ k_0$. So for $k ≥ k_0$ we have $|x_k|^q < |x_k|^p$ which implies ...

1

If $f$ is known to be an involution, all you need to prove that it's an isometry is the Lipschitz property $d(f(x),f(y))\le d(x,y)$ for all $x,y$. Since $f\circ f=\textrm{id}$, it then follows that $d(f(x),f(y)) = d(x,y)$ for all $x,y$, and that $f$ is onto. I somehow don't see the connection of the above with the second question; the symmetry ...

5

Hint: use the Holder inequality.

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