Proving not so usual topology for $C(I)= \lbrace f:I \to \mathbb{R}:\mbox{f is continuous} \rbrace$ using the definition of a basis.

Question

Let $I=[0,1]$ and $C(I)= \lbrace f:I \to \mathbb{R}:\mbox{f is continuous} \rbrace$. Also for every $f \in C(I)$ , every $n \in \mathbb{N}$, every collection $\lbrace x_{1},....,x_{n} \rbrace \subset I$ and every $r>0$, we have the collection

$$[f;x_{1},...,x_{n},r]=\lbrace g \in C(I): \mbox{for every }\: 1\leq i \leq n\:(|f(x_{i}-g(x_{i}))|<r) \rbrace$$.

Also consider the set

$$B_{p}= \lbrace [f;x_{1},...,x_{n},r]: f \in C(I),\: n \in \mathbb{N}, \:x_{1},....,x_{n} \in I, \:r>0 \rbrace$$

I want to prove the following family of subset is a topology for $C(I)$:

$$\tau_{p}= \lbrace E \subset C(I):\mbox{there is }\:\mathcal{A} \subset B_{p}\: \mbox{with}\: E= \bigcup \mathcal{A} \rbrace$$.

So in order to prove $(C(I), \tau_{p})$ is a topological space I prove that $B_{p}$ is basis for $T_{p}$ as the definition of basis here:

https://en.wikipedia.org/wiki/Base_(topology)

(1) No problem showing that $C(I)= B_{p}$.

My attempt to prove (2) goes as follows:

(2) Let $$ h \in [f;x_{1},...,x_{n},r] \cap [g;y_{1},...,y_{m},s] $$.

As $h \in [f;x_{1},...,x_{n},r]$, then I take $\epsilon'=max \lbrace|f(x_{i}-h(x_{i}))|<r : i=1,...,n \rbrace$. And as $h \in [g;y_{1},...,y_{m},s]$, then I take $\delta'=max \lbrace|g(y_{i}-h(y_{i}))|<s : i=1,...,m \rbrace$. And I define $\epsilon=r-\epsilon'$ and $\delta= r- \delta'$. As $\epsilon, \delta >0$ I take $t>0$ such $0< t<min(\epsilon,\delta)$. So I need to prove that

$$ [h;x_{1},...,x_{n},y_{1},...,y_{m},t] \in [f;x_{1},...,x_{n},r] \cap [g;y_{1},...,y_{m},s] $$.

So I take $p \in [h;x_{1},...,x_{n},y_{1},...,y_{m},t]$ but I cannot prove this $p$ is the intersection above.

Answer 1

We have some $h \in [f;x_{1},...,x_{n},r] \cap [g;y_{1},...,y_{m},s]$. You have to choose sliightly different $\delta$'s per coordinate:

Some notational stuff: Let $I_1= \{x_1,x_2,\ldots,x_n\}$ and $I_2=\{y_1,y_2,\ldots, y_m\}$. (We could denote the first open set by $[f;I_1,r]$ and the second by $[g;I_2,s]$, note that the order of the points is irrelevant).

Now $I_1 \cup I_2 = (I_1 \cap I_2) \cup (I_1 \Delta I_2)$, where the second set is the symmetric difference of the sets, so the points where we only have one condition to contend with, vs. $I_1 \cap I_2$ (could be empty) where we have two conditions.

For every $z \in I_1 \cap I_2$ we thus have that $h(z) \in B(f(z), r) \cap B(f(z),s)$ (where $B(a,\delta)=\{a' \in \Bbb R: |a'-a| < \delta\}$ is the usual open ball (and open interval here) in $\Bbb R$) and as these balls are open (and finite intersections also), we can find $\delta_z >0$ such that $B(z,\delta_z) \subseteq B(f(z), r) \cap B(f(z),s)$ (we can find an explicit $\delta_z$ by $\min(r-|f(z)-h(z)|, s-|f(z)-h(z)|)$, but it's enough to know that such a $\delta_z$ exists, I don't want to check here that it works).

For every $z \in I_1 \Delta I_2$ we just have one condition: either $h(z) \in B(f(z),r)$ or $h(z) \in B(g(z), s)$ and in either case we find another $\delta_z>0$ such that either $B(h(z), \delta_z) \subseteq B(f(z), r)$ (e.g. $\delta_x = r-|h(z)-f(z)|$ etc.) or $B(h(z), \delta_z) \subseteq B(g(z), s)$ (etc.).

Finally, set $\delta = \min\{\delta_z: z \in I_1 \cup I_2\} >0$ and note that now by construction:

$$h \in [h; I_1\cup I_2; \delta] \subseteq [f;x_{1},...,x_{n},r] \cap [g;y_{1},...,y_{m},s]$$

Note that I just used that all sets $B(f(z),r)$ and $B(g(z),s)$ are open in $\Bbb R$ and no more, really.