Unpacking a proof that filter convergence characterizes continuous maps Statement:

Let $f : (X, \tau) \to (Y, \sigma)$ be a mapping between topological spaces. Then $f$ is continuous iff, whenever $\Phi$ is a filterbase on $X$ converging to a limit $p \in X$, then the filterbase $f(\Phi)$ converges to $f(p) \in (Y, \sigma)$.

Proof:

(i) Suppose $f$ is continuous. Let the filterbase $\Phi$ converge to $p$ (that is, each neighbourhood of $p$ contains some $F \in \Phi$). If $N$ is a neighbourhood of $f(p) \in Y$ then $f^{–1}(N)$ is$_1$ a neighbourhood of $p \in X$ so $f^{–1}(N)$ contains some $F \in \Phi$, that is, $f(F) \subseteq_2 N$; therefore $f(\Phi)$ converges$_3$ to $f(p)$.


(ii) Suppose $f$ is not continuous. Then $\exists$ closed $K \subseteq Y$ such that $f ^{–1}(K)$ is not closed in $X$, so $\exists \ p \in \overline{f^{–1}(K)}$ such that $p \not \in_4 f^{–1}(K)$. Then $\Phi = \{N \cap f^{–1}(K) : N \in \text{set of neighborhoods of $p$}\}$ is$_5$ a filterbase that converges$_6$ to $p$. Yet $f(\Phi)$ does not$_7$ converge to $f(p)$, since $Y \setminus K$ is a neighbourhood of $f(p)$ containing no $f(F)$ for $F \in \Phi$.

My questions (each index in the proof above corresponds to the number of a question below):

*

*Suppose $N$ is a neighborhood of $f(p).$ Then by definition of neighborhood,  there some open $G \in \sigma$ s.t. $f(p) \in G \subseteq N$. Then $p \in f^{-1}(G).$ But $G \subseteq N \implies f^{-1}(G) \subseteq f^{-1}(N)$. By definition, $f^{-1}(N)$ is a neighborhood of $p$. Is that correct?


*(a) Suppose $y \in f(F).$ Then there is $p \in X$ s.t. $f(p) = y.$ And so there's some $F \subseteq X$ s.t. $p \in F$. By assumption, each neighbourhood of $p$ contains some $F \in \Phi$ and since $f^{-1}(N)$ is a neighborhood of $p$, we have $F \subseteq f^{-1}(N)$ meaning $p \in f^{-1}(N)$ implying $f(p) = y \in N.$ Is that correct?
(b) Why can't we simply say $f(p) \in f(F) \implies f(p) \in N$ as $N$ is a neighborhood of $f(p)$?


*That's true because $\Lambda = \{N \subseteq Y: \exists f(F) \in f(\Phi), \ f(F) \subseteq N\}$ is a filter on $Y$ and $f(\Phi)$ is a base for $\Lambda.$ Since $\Lambda \to f(p)$, we have $f(\Phi) \to f(p)$ by definition. Is that correct?


*Since $f^{-1}(K)$ is not closed, $f^{-1}(K) \ne \overline{f^{-1}(K)}$?


*$\Phi$ is a filterbase because $\{p\} \subseteq \left((N_1) \cap f^{-1}(K)\right) \cap \left((N_2) \cap f^{-1}(K)\right)$ (is this correct?) and $\varnothing \not \in \Phi.$ But how do we know the latter? What if $f^{-1}(K) = \emptyset?$ Then $N \cap f^{-1}(K) = \emptyset?$


*$\Phi$ converges to $p$ because $N \cap f^{-1}(K) \subseteq N$ and the set of neighborhoods of $p$ - $\mathcal N(p)$ - is a filter on $X$ that converges to $p$ and $\Phi$ is a base for $\mathcal N(p)$. Is that correct?


*Suppose $y \in f(F).$ Then $y \in f(N \cap f^{-1}(K))$. And so there's some $p \in X$ s.t. $f(p) = y.$ That means we can find some $N_1$ s.t. $N_1 \cap f^{-1}(K) \subseteq X$ with $p \in N_1 \cap f^{-1}(K)$ meaning $p \in f^{-1}(K)$ impliying $f(p) \in K$ further implying $f(p) \not \in Y \setminus K.$ Thus $f(F) \not \subseteq Y \setminus K.$ Correct?
 A: *

*Yes that is correct.

*$F \subseteq f^{-1}[N]$ implies that any $x \in F$ is mapped under $f$ into $N$, so $f[F] \subseteq N$ is immediate. Your argument is convoluted IMO.

*is clear because we just showed that any neighbourhood of $f(p)$ contains some member of $f(\Phi)$ and that is the definition of convergence of filterbases that you just yourself stated. So no further argument is needed. Lose the $\Lambda$; it's not needed.

*Yes, that is the reason. $f^{-1}[K]$ is not equal to its closure, so the closure is a strict superset, which is exactly what the statement says.

*$\Phi$ is a filterbase, because $p \in \overline{f^{-1}[K]}$ tells us that all sets $N \cap f^{-1}[K]$, where $N$ ranges over all neighbourhoods of $p$, are non-empty. The set $\Phi$ is obviously closed under intersections as $N_1 \cap f^{-1}[K]) \cap (N_2 \cap f^{-1}[K]) = (N_1 \cap N_2) \cap f^{-1}[K]$ where the latter is in $\Phi$ again, as $N_1 \cap N_2$ is surely still a neighbourhood of $p$ etc. So the conditions on being a filterbase are trivially satisfied.

*is correct: any neighbourhood $N$ of $p$ contains "its own" filterbase member $N \cap f^{-1}[K]$ trivially so convergence of $\Phi$ to $p$ is immediate by the definition of convergence of filterbases.

*The set $F:=N \cap f^{-1}[K]$ (a typical member of $\Phi$) is mapped under $f$ into $K$ because $F$ is a subset of $f^{-1}[K]$ in particular. It follows immediately that $f[F] \subseteq K$ or equivalently $f[F] \cap (Y\setminus K) = \emptyset$ and the claim has been shown. You waste too many words in your version of this argument. It's really this simple.

