Originally Posted by
Opalg It's not a metric topology, basically because there are uncountably many of the defining seminorms $\displaystyle \rho_x$. A basic family of neighbourhoods of f would consist of sets of the form $\displaystyle \{g\in\mathbb{X}:\rho_{x_j}(f-g) = |f(x_j)-g(x_j)|<\varepsilon\; (1\leqslant j\leqslant n)\}$ for each finite set of points $\displaystyle x_1,...,x_n\in[0,1].$ It is called the topology of pointwise convergence because a sequence (or more generally a directed net) $\displaystyle \{f_i\}$ converges to the limit $\displaystyle f$ iff $\displaystyle f_i(x)\to f(x)$ for each point $\displaystyle x\in[0,1].$
For the last part of the question, I think that you have to use a cardinality argument. You want to find a sequence $\displaystyle (f_n)$ in $\displaystyle \mathbb{X}$ which converges to 0 (in the topology of pointwise convergence) so slowly that if you multiply its terms by any sequence of scalars that tends to infinity then the resulting sequence $\displaystyle (\gamma_nf_n)$ no longer converges to 0.
The set $\displaystyle \Gamma$ of all sequences of integers that tend to infinity has cardinality $\displaystyle 2^{\aleph_0}$, which is the same as the cardinality of [0,1]. So there exists a bijective map $\displaystyle \phi:\Gamma\to [0,1]$. Define the function $\displaystyle f_n\in\mathbb{X}$ by $\displaystyle f_n(\phi(\gamma)) = 1/\gamma_n\ (\gamma\in\Gamma)$.
Then $\displaystyle f_n(x)\stackrel{n}{\to}0$ for all $\displaystyle x\in[0,1]$, but $\displaystyle \gamma_nf_n(\phi(\gamma)) = 1\stackrel{n}{\not\to}0.$
Finally, I leave you to check that if $\displaystyle (\gamma_nf_n)$ fails to converge to 0 for all sequences of integers tending to infinity, then the same will be true for all sequences of reals tending to infinity.