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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 $B$.

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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 $i\geq i_0$.

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  • $\begingroup$ 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 $\tau_1=\tau_2$? $\endgroup$
    – 0xbadf00d
    Commented Dec 22, 2020 at 14:28
  • $\begingroup$ Yes, this is exactly what I mean. $\endgroup$
    – Ruy
    Commented Dec 22, 2020 at 14:30
  • $\begingroup$ 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 $\tau_1=\tau_2$? $\endgroup$
    – 0xbadf00d
    Commented Dec 22, 2020 at 14:30
  • $\begingroup$ (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) $\endgroup$
    – 0xbadf00d
    Commented Dec 22, 2020 at 14:31
  • $\begingroup$ 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 $x$. $\endgroup$
    – 0xbadf00d
    Commented Dec 22, 2020 at 14:32

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