Semi-norms in Functional Analysis I'm self-studying functional analysis. The following is from Rudin's "Functional Analysis, 2nd edition". It consists of parts from question 7 and 13 from the first chapter. I am not sure if my answers are correct, and would appreciate any hints/help. Thanks!
7) Let $X$ be the vector space of all complex functions on the unit interval $[0,1]$, topologized by the family of semi-norms
\begin{equation}
p_x(f) = |f(x)|\,\,\,\,\,\, 0\leq x\leq 1
\end{equation}
Show that there is a sequence $\{f_n\}$ in $X$ such that $\{f_n\}$ converges to $0$ as $n\rightarrow \infty$, but if $\{\gamma_n\}$ is any sequence of scalars such that $\gamma_n \rightarrow \infty$, then $\{\gamma_nf_n\}$ does not converge to $0$ (Hint: use the fact that the collection of all complex sequences converging to $0$ has the same cardinality as $[0,1]$.)
13) Let $C$ be the vector space of all complex continuous functions on $[0,1]$. Define
\begin{equation}
d(f,g) = \int_0^1\frac{|f(x) - g(x)|}{1+|f(x) - g(x)|}dx
\end{equation}
Let $(C,\sigma)$ be $C$ with the topology induced by this metric. Let $(C,\tau)$ be the topological vector space defined by the semi-norms
\begin{equation}
p_x(f) = |f(x)|\,\,\,\,\,0\leq x\leq1
\end{equation}
a) Prove that every $\tau$-bounded set in $C$ is also $\sigma$-bounded and that the identity map $id:(C,\tau)\rightarrow (C,\sigma)$ therefore carries bounded sets into bounded sets.
b)Prove that $id:(C,\tau)\rightarrow (C,\sigma)$ is not continuous, although it is sequentially continuous. Hence $(C,\tau)$ is not metrizable. Show also directly that $(C,\tau)$ has no countable local base.
Attempt:
7) Im really struggling with this one. I don't see where the cardinality of $[0,1]$ comes in. 
13) a) Let $E$ be a $\tau$-bounded set in $C$. Using Theorem 1.37 from the book, we can say that in the $\tau$-topology, a set $E\subset C$ is bounded if and only if every $p_x(f)$ is bounded on $E$. i.e., for any $f\in E$, $|f(x)| < k_x$ for all $x\in [0,1]$, where $k_x$ is some positive real number dependent on $x$.
If $g\in E$ then $d(g,0)<1$, because $|g(x)|<\infty$, from the above statement. Let $V_{\sigma} = \{g\in C : d(g,0) < 1\}$. Then, $tV_{\sigma} = \{f\in C : d(f,0)<t\}$. Therefore, if $h\in E$, then $h\in tV_{\sigma}$ for all $t\geq s$, for $s$ large enough, and so $E$ is also $\sigma$-bounded.
b) If $\{f_n\}\rightarrow f$ in the $\tau$-topology, then $d(f_n,f)\rightarrow 0$, and so $id$ is sequentially continuous.
To show $(C,\tau)$ does not have a countable base: Let $\mathcal{B}$ be the base constructed by the family of semi-norms. i.e., $\mathcal{B}$ is the collection of all finite intersections of sets $V(p_x,n) = \{f:p_x(f)<\frac{1}{n}\}$, $n$ a positive integer. For a real number $x$ and an integer $n$, a base element is given by:
\begin{equation}
B_{x,n} = \{f\in C:|f(x)|<\frac{1}{n}\}
\end{equation}
Because $[0,1]$ is uncountable, there is an uncountable number of base elements. We can therefore conclude that $id$ is not continuous (sequential continuity result, found in A6 in appendix of the book), and that $(C,\tau)$ is not metrizable.
 A: Hint for exercise 7: Let $\mathcal{Z}$ be the set of complex sequences converging to $0$. Let $b\colon \mathcal{Z} \to [0,1]$ be a bijection. What might be a good choice for the value of $f_n(b(\zeta))$ then?

 We choose $f_n(b(\zeta)) = \zeta_n$. If $(\gamma_n)$ is a sequence with $\gamma_n \to \infty$, define $\hat\zeta$ by $\hat\zeta_n = 1/\gamma_n$ if $\gamma_n\neq 0$, and $\hat\zeta_n = 0$ if $\gamma_n = 0$. Then $\gamma_n\cdot f_n(b(\hat\zeta)) = 1$ for all large enough $n$.

Concerning 13 a): No, the distance $d$ is not homogeneous,
$$t\cdot V_\sigma \neq \left\{ f\in C : d(f,0) < t\right\}.$$
If you define $\varphi(x) = \sup \{ \lvert f(x)\rvert : f \in B\}$ for a $\tau$-bounded subset $B\subset C$, what regularity properties of $\varphi$ do you know? And what has Lebesgue to say about the sequence $$\frac{\varphi/n}{1+\varphi/n}\,?$$

 $\varphi$ is the supremum of continuous functions, hence it is lower semicontinuous, and therefore measurable. Also, $\varphi$ is non-negative and finite everywhere, since $B$ is $\tau$-bounded, so the sequence $\psi_n = \frac{\varphi/n}{1+\varphi/n} = \frac{\varphi}{n+\varphi}$ of measurable functions converges to $0$ pointwise, and it is evidently dominated by the constant function $1$. By Lebesgue's dominated convergence theorem, or by the monotone convergence theorem, it follows that $\int_0^1 \psi_n(x)\,dx \to 0$. Hence for every $r > 0$, there is an $n_r\in \mathbb{N}\setminus \{0\}$ such that $0 \leqslant \int_0^1 \psi_n(x)\,dx < r$ for all $n \geqslant n_r$. But that implies $n^{-1}\cdot B \subset B_r(0)$ for all $n\geqslant n_r$, so $B$ is shown to be $\sigma$-bounded.

Concerning 13 b): What is your argument for $d(f_n,f) \to 0$ when $f_n \to f$ pointwise?

 The dominated convergence theorem. $0 \leqslant \delta_n(x) := \frac{\lvert f_n(x) - f(x)\rvert}{1+\lvert f_n(x) - f(x)\rvert} \leqslant 1$ for all $n$ and $x$, and $\delta_n \to 0$ pointwise. Thus the identity is sequentially continuous (which follows from the fact that it maps $\tau$-bounded sets to $\sigma$-bounded sets).

Why is $\operatorname{id} \colon (C,\tau) \to (C,\sigma)$ not continuous, although it is sequentially continuous?

 Because not all $\sigma$-neighbourhoods of $0$ are $\tau$-neighbourhoods of $0$, see next point.

Can you find a neighbourhood $V$ of $0$ in $(C,\sigma)$ such that no $\tau$-neighbourhood of $0$ is contained in $V$?

 Any $d$-ball $B_r(0)$ with $0 < r < 1$ will do. Since every $\tau$-neighbourhood of $0$ only constrains the values at finitely many points, every $\tau$-neighbourhood of $0$ contains functions with $d(f,0)$ arbitrarily close to $1$. Take a function that vanishes at the finitely many points controlled by the given $\tau$-neighbourhood, and has sufficiently high triangular spikes in between.

To show that $\tau$ has no countable local basis at $0$, it is not sufficient to expose one local basis that is uncountable. You must show that no countable system can be a neighbourhood basis of $0$.
How does a neighbourhood of $0$ in $\tau$ look, and hence, what can you say about a countable family of neighbourhoods? Which points of $[0,1]$ are controlled by any of these neighbourhoods?

 A $\tau$-neighbourhood of $0$ is a set containing $T(\varepsilon; x_1,\dotsc,x_k) = \{ f \in C : \lvert f(x_i)\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant k\}$ for some $\varepsilon > 0$, non-negative integer $k$ and points $x_1,\dotsc,x_k \in [0,1]$. So a countable family $\mathscr{U} = \{U_n : n \in \mathbb{N}\}$ of neighbourhoods ($U_n \supset T(\varepsilon_n;x_1^n,\dotsc x_{k_n}^n)$) only constrains the values at countably many points. For any $x\in [0,1]\setminus \bigcup_n \{x_i^n : 1 \leqslant i \leqslant k_n\}$, none of the $U_n$ is contained in $T(1; x)$, so $\mathscr{U}$ is not a local base. Since $\mathscr{U}$ was an arbitrary countable family of neighbourhoods, it follows that any local base must be uncountable.

