If $Σ$ is endowed with the initial topology $\cal T_ Φ$ corresponding to the collection $Φ$ then $σ_i⟶σ$ iff $φ(σ_i)⟶φ(σ)$ for any $φ∈Φ$.

Question

When I started to study topology of compact convergence it seemed to me that there was something not said explicitly that really help to understand better this topology so that I observed that for any metri space $(Y,d)$ and for any topological space $(X,\cal T)$ if $Y^X$ is equipped with the compact topology $\mathcal T_C$ corresponding to $d$ then any projection $\pi_K$ of $Y^X$ onto $Y^K$ with $K\in\mathcal P(X)$ compact is continuous when $Y^K$ is equipped with topology of uniform convergence $\mathcal T_\infty(Y^K)$ and really this is true since actually $\mathcal T_C$ is the initial topology corresponding to these projections as here is proved. Anyway, when Munkres into his Topology text started to explain the topology of compact convergence then he gives as exercise to show that a sequence of functions $(f_n)_{n\in\omega}$ in $Y^X$ converges to any function $f$ if and only if for any compact $K\in\mathcal P(X)$ the restricted sequence $\big(f_n|_K\big)_{n\in\omega}$ converges to $f|_K$ so that I observed that this was trivial if the following proposition

Proposition 1

If $\Sigma$ is a set equipped with the initial topology $\cal T_\Phi$ corresponding to the collection of functions $\Phi$ then a net $(\sigma_i)_{i\in I}$ in $\Sigma$ based on the directed set $(I,\preceq_I)$ converges to an $\sigma\in\Sigma$ if and only if for any $\varphi\in\Phi$ the net $\big(\varphi(\sigma_i)\big)_{i\in I}$ converges to $\varphi(\sigma)$

holds.

So, I observed that if $(\sigma_i)_{i\in I}$ converges to any $\sigma$ then by continuity of $\varphi\in\Phi$ the net $\big(\varphi(\sigma_i)\big)_{i\in I}$ converges to $\varphi(\sigma)$. However, for any $\varphi\in\Phi$ let be $\Lambda_\varphi$ the codomain of $\varphi$ and so $\cal T_\varphi$ the corresponding topology for $\Lambda_\varphi$ so that if $A_\sigma\in\cal T_\Phi$ is a(n) (open) neighborhood of $\sigma$ then there exists $(\boldsymbol\varphi,\boldsymbol A)\in\prod_{p\in q}\Phi\times\prod_{p\in q}\cal T_{\varphi_p}$ with $q\in\omega$ such that
$$
\sigma\in\bigcap_{p\in q}\varphi^{-1}_p[A_p]\subseteq A_\sigma
$$
where we put
$$
\varphi_p:=\boldsymbol\varphi(p)\quad\text{and}\quad A_p:=\boldsymbol A(p)
$$
for any $p\in q$. So if $\big(\varphi(\sigma_i)\big)_{i\in I}$ converges to $\varphi(\sigma)$ for each $\varphi\in\Phi$ then for any $p\in q$ there exists $i_p\in I$ such that
$$
\varphi(\sigma_i)\in A_p
$$
for any $i\succeq_I i_p$ so that fixing by cofinality $i_0\in I$ such that
$$
i_p\preceq_I i_0
$$
for any $p\in q$ then for any $i\succeq_I i_0$ we have
$$
\sigma_i\in\bigcap_{p\in q}\varphi^{-1}_p[A_p]\subseteq A_\sigma
$$
which proves that $(\sigma_i)_{i\in I}$ converges to $\sigma$.

So, I realise that what I did can appear a not hard result so that the question can be reputed unnecessary but I decided to pose it because many (notable) authors (e.g. Engelking, Willard, Kelley etc…) prove proposition $1$ only when $\Sigma$ is just a product of any collection $\frak S$ and this led me to think that it could not be true if this is not. So, could someone help me, please?

Answer 1

The proof of Proposition 1 contained in your question is correct. In your wiki-answer you prove more, namely that the initial topology is the unique topology on $\Sigma$ with the property

A net $(\sigma)_{i\in I}$ in $\Sigma$ based on the directed set $(I,\preceq_I)$ converges to any $\sigma$ if and only if the net $\big(\varphi(\sigma_i)\big)_{i\in I}$ converges to $\varphi(\sigma)$ for any $\varphi\in\Phi$.

Your proof is correct, but there is an alternative and more general theorem covering uniqueness.

Theorem. A topology on a set $X$ is uniquely determined by its convergent nets. More precisely, the closed subsets of $X$ are characterized as follows:
$A \subset X$ is closed if and only for each net $(x_i)_{i\in I}$ with $x_i \in A$ and $x_i \to x \in X$ one has $x \in A$.

Proof. Let $A \subset X$ be closed and $(x_i)_{i\in I}$ be a net with $x_i \in A$ and $x_i \to x \in X$. Assume $x \notin A$. Then $U = X \setminus A$ is an open neigborhood of $x$ which must contain some $x_i$ because $x_i \to x$. This is a contradiction, therefore $x \in A$.

To prove the converse assume that for each net $(x_i)_{i\in I}$ with $x_i \in A$ and $x_i \to x \in X$ one has $x \in A$. Now consider $x \in \overline A$ . This means that each open neighborhood $U$ of $x$ contains a point $x_U \in A$. The set $\mathcal U(x)$ of all open neighborhoods if $x$ in $X$ can be partially ordered via $U \preceq V$ if $V \subset U$. This makes $(\mathcal U(x),\preceq )$ a directed set and $(x_U)$ a net with $x_U \in A$. Then $x_U \to x$ because for each $U \in \mathcal U(x)$ one has $x_V \in V \subset U$ for all $V \succeq U$. Therefore $x \in A$.