Let $F([0,1], \mathbb{R})$ the space of functions of [0,1] in $\mathbb{R}$, together with the topology of the point convergence. Let $F([0,1], \mathbb{R})$ the space of functions of [0,1] in $\mathbb{R}$, together with the topology of the point convergence $\tau$.
Show that the set of continuous functions of [0,1] in $\mathbb{R}$ is dense in $(F([0,1], \mathbb{R}), \tau)$.
 A: let $U$ be an open of $(F (([0,1]), $\mathbb{R}$), T)$ such that $ f \in U $. by the point convergence topology there are $a_1, ..., a_n$ in $[0,1]$ and $W_ {a_i}$ open in $\mathbb{R}$ for each $ i = 1,2, .., n $ such that
$f \in \cap_{i=1}^{n} \pi_{a_i}^{-1}(W_{a_i}) \in U$.
A: Lemma:

If $X$ is a Tychonoff space and $\{x_1,\ldots, x_n\}$ is a finite subset of $X$, and $y_1, \ldots, y_n \in \Bbb R$, then there exists a continuous function $f:X \to \Bbb R$ such that $f(x_i) = y_i$ for all $i$.

I'll do the easy special case of $X=[0,1]$, that we need here. We can just use polynomial interpolation to find an $(n-1)$-degree polynomial $p: \Bbb R \to \Bbb R$ such that $p(x_i)=y_i$ and polynomials are continuous (and much more) and then $p\restriction_{[0,1]}$ is as required. We can also use a piecewise linear function, consisting of different linear functions $f(t)=at+b$ on the different intervals $[0,x_1], [x_1,x_2], \ldots [x_n,1]$ (when we order $0 \le x_1 < x_2 <\ldots x_n \le 1$), which might even be easier. But we can find such continuous for all Tychonoff (completely regular $T_1$) spaces in general, with just a little more work.
Now, if $U=\bigcap_{i=1}^n \pi_{x_i}^{-1}[W_{x_i}]$ is a basic non-empty open subset of the pointwise topology on $\Bbb R^{[0,1]}$, we pick $y_i \in W_{x_i}$ for all $i$ and apply the lemma to get $f: [0,1]\to \Bbb R$ continuous such that $f(x_i)=y_i$. But this exactly means that for all $i$, $$\pi_{x_i}(f)=f(x_i)=y_i \in W_{x_i} \implies f \in \pi_{x_i}^{-1}[W_{x_i}]$$
and so $f \in U$ and so the continuous functions are dense.
The crux is that any finite set of conditions on a function (which is what $U$ is, really) can be fulfilled by a continuous function too, because we have so many of them.
