# Do the topologies of pointwise and compact convergence coincide on equicontinuous subsets?

If $$E$$ is a set, then the topology $$\rho(E)$$ generated by $$p_x(f):=|f(x)|\;\;\;\text{for }f:E\to\mathbb R$$ for $$x\in E$$ is called the topology of pointwise convergence on $$\mathbb R^E$$. If $$\tau$$ is a topology on $$E$$, then the topology $$\kappa(E,\tau)$$ generated by $$p_K(f):=\sup_{x\in K}|f(x)|\;\;\;\text{for }f\in C(E,\tau)$$ for $$\tau$$-compact $$K\subseteq E$$ is called the topology of compact convergence on $$C(E,\tau)$$.

Are we able to show that if $$\Gamma\subseteq C(E,\tau)$$ is $$\tau$$-equicontinuous, then$$^1$$ $$\left.\rho(E)\right|_\Gamma=\left.\kappa(E,\tau)\right|_\Gamma$$? If not, what do we need to assume to show that?

Moreover, I would like to know how exactly we can show that $$\kappa(E,\tau)\subseteq\left.\rho(E)\right|_{C(E,\:\tau)}\tag1.$$

Remark: I know that generally, if $$X$$ is a vector space and $$\sigma$$ is the topology generated by a family $$P$$ of seminorms on $$X$$, then $$\mathcal B_P:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}$$ is an analytic basis for $$\sigma$$, where $$U_p:=\{x\in X:p(x)<1\}\;\;\;\text{for }p\in P.$$ This should be helpful to know.

$$^1$$ If $$(X,\sigma)$$ is a topological space and $$B\subseteq X$$, then $$\left.\sigma\right|_B:=\{O\cap B:O\in\sigma\}$$ denotes the subspace topology on $$B$$.

Let $$X$$ be a topological space, $$M$$ be a metric space, and $$E$$ be an equicontinuous set of functions from $$X$$ to $$M$$. Also let $$D$$ be a dense subset of $$X$$. Then the following three topologies coincide on $$E$$:

1. The topology of uniform convergence on compact subsets of $$X$$,
2. The topology of pointwise convergence on $$X$$,
3. The topology of pointwise convergence on $$D$$. In order to verify this assertion it suffices to prove that if $$\{f_i\}_{i\in I}$$ is a net in $$E$$, and if $$f\in E$$, then the following are equivalent:

a. $$\displaystyle\lim_{i\to \infty }\sup_{x\in K}d\big (f_i(x), f(x)\big )=0$$, for every compact subset $$K\subseteq X$$,

b. $$\displaystyle\lim_{i\to \infty }f_i(x)=f(x)$$, for every $$x\in X$$,

c. $$\displaystyle\lim_{i\to \infty }f_i(x)=f(x)$$, for every $$x\in D$$.

It is evident that (a) $$\Rightarrow$$ (b) $$\Rightarrow$$ (c), so we only need to worry about (c) $$\Rightarrow$$ (a).

In order to do this, given any compact subset $$K\subseteq X$$, and any $$\varepsilon >0$$, using equicontinuity, for each $$x$$ in $$K$$, we choose a neighborhood $$V_x$$ of $$x$$ such that $$y\in V_x\Rightarrow d\big (g(x),g(y)\big )<\varepsilon, \quad\forall g\in E.$$ Since $$K$$ is compact and $$\{V_x\}_{x\in K}$$ is an open cover of $$K$$, we may find $$x_1,x_2,\ldots ,x_n\in K$$ such that $$K\subseteq \bigcup_{j=1}^nV_{x_j}.$$ Since $$D$$ is dense in $$X$$, for every $$j$$ we may pick some $$y_j\in D\cap V_{x_j}$$. Using (c) let $$i_0\in I$$, such that $$i\geq i_0\Rightarrow d\big (f_i(y_j), f(y_j)\big )<\varepsilon , \quad\forall j=1,\ldots ,n.$$ Next, given any $$i\geq i_0$$, and any $$x$$ in $$K$$, pick $$j$$ such that $$x\in V_{x_j}$$. Then $$d\big (f_i(x),f(x)\big ) \leq$$$$\leq d\big (f_i(x),f_i(x_j)\big ) + d\big (f_i(x_j),f_i(y_j)\big ) + d\big (f_i(y_j), f(y_j)\big ) +$$$$+ d\big (f(y_j), f(x_j) \big ) + d\big (f(x_j), f(x) \big ) < 5\varepsilon .$$ Consequently $$\sup_{x\in K}d\big (f_i(x), f(x)\big )\leq 5\varepsilon ,$$ for all $$i\geq i_0$$.

• When you write "it suffices to prove", do you've got the following in mind: If $E$ is a set and $\tau_i$ is a topology on $E$ such that every net has a $\tau_1$-limit point if and only if it has a $\tau_2$-limit, then $\tau_1=\tau_2$? Commented Dec 22, 2020 at 14:28
• Yes, this is exactly what I mean.
– Ruy
Commented Dec 22, 2020 at 14:30
• I'm actually not 100% sure whether this is correct or it needs to be stated in the following way instead: If $E$ is a set and $\tau_i$ is a topology on $E$ such that for every net $(x_t)_{t\in I}\subseteq E$ and $x\in E$ it holds that $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$ if and only if $x$ is a $\tau_2$-limit of $(x_t)_{t\in I}$, then $\tau_1=\tau_2$? Commented Dec 22, 2020 at 14:30
• (The difference is that in the former characterization, it might be that the $\tau_1$-limit point and the $\tau_2$-limit may be different) Commented Dec 22, 2020 at 14:31
• I'm sure that in either case, we would derive this characterization from the fact that if $B\subseteq E$ and $x\in E$, then $x\in\overline B$ if and only if there is a net converging to $x$. Commented Dec 22, 2020 at 14:32