Prove continuity from preservation of sequential limits but do not use $f(\bar{A}) \subseteq \overline{f(A)}$

Question

Munkres Topology:

Theorem 21.3. Let $f : X \rightarrow Y$. If the function $f$ is continuous, then for every convergent sequence $x_n \rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is metrizable.

I understand the first statement. My question is about the converse assuming that $X$ is metrizable. Munkres does this by proving the equivalent formulation of continuity $f(\bar{A}) \subseteq \overline{f(A)}, A \subseteq X$. How do I prove that $f$ directly satisfies the original definition of continuity?

Definition of continuity: For each open subset V of Y, the set $f^{-1}(V)$ is an open subset of X.

I can type up my proof later (I have to organize my messy notes), but for now: I was able to show that $f^{-1}(V)$ is a subset of $f^{-1}(V)'$, which I discovered to be something called dense-in-itself which means no isolated points.

I think I can show $f^{-1}(V)$ is open by showing that $f^{-1}(V)' \cap f^{-1}(V^c)'$ is empty because

• $f^{-1}(V^c) = f^{-1}(V)^c$,

• $U$ is open if and only if $Bd(U)=\overline{U} \setminus U$

• showing that $f^{-1}(V)' \cap f^{-1}(V^c)'$ is empty would imply $Bd(U)=\overline{U} \setminus U$ for $U=f^{-1}(V)$ because $f^{-1}(V) \subseteq f^{-1}(V)'$

What is the contradiction if we say that $f^{-1}(V)' \cap f^{-1}(V^c)'$ is not empty? I'm very confused as to what we get from assuming there is some $z \in f^{-1}(V)'$ and $z \in f^{-1}(V^c)'$.

From $z \in f^{-1}(V)'$, I think I was able to show that $(f^{-1}(V)')'=f^{-1}(V)'$, which means that $f^{-1}(V^c)'$ is something I discovered to be called a perfect set, which means closed and dense-in-itself. According to proofwiki.org, the Union of Set of Dense-in-itself Sets is Dense-in-itself. This means $f^{-1}(V) \cup f^{-1}(V)'$ is dense-in-itself, but $f^{-1}(V) \cup f^{-1}(V)' = \overline{f^{-1}(V)}$, the closure of $f^{-1}(V)$ which is closed. Therefore, $\overline{f^{-1}(V)}$ is perfect.

• Is this related? Does continuity of $f$ imply $f^{-1}(\bar A)\subset\overline{f^{-1}(A)}$?

I do not know how to proceed from $z \in f^{-1}(V^c)'$. I think this somehow contradicts with $\overline{f^{-1}(V)}$ being perfect.

Also, I showed $f^{-1}(V)$ is infinite if we let $x \in f^{-1}(V)'$ because the sequence lemma gives us a sequence $x_n \to x$ where $x_n \in f^{-1}(V)$ . Also, $f(x_n) \in V$ and $f \in \overline{V}$.

However, $f^{-1}(V)$ is infinite, $\overline{f^{-1}(V)}$ is dense-in-itself and $(f^{-1}(V)')' \subseteq f^{-1}(V)'$ can be shown from $T_1$ axiom because according to proofwiki.org again, Dense-in-itself Subset of T1 Space is Infinite, Closure of Dense-in-itself is Dense-in-itself in T1 Space and Derivative of Derivative is Subset of Derivative in T1 Space. Thus, I think there's something else I'm not using. (Of course, I'm not using $f(\bar{A}) \subseteq \overline{f(A)}$ because that's what this question is about.)

Answer 1

The argument should be elementary.

Take $x\in X$ and a sequence $x_n\in X$ with $x_n\to x$. We want to prove that $f(x_n)$ converges to $f(x)$, so take an open neighborhood $V$ of $f(x)$. By continuity, $f^{-1}(V)$ is open, and it clearly contains $x$. Since $x_n\to x$, there exists $N\in\mathbb{N}$ such that $x_n\in f^{-1}(V)$ for $n>N$. Then, for $n>N$, it is true that $f(x_n)\in f(f^{-1}(V))\subseteq V$.

We proved: for every open set $V$ with $f(x)\in V$ there exists $N\in\mathbb{N}$ such that $f(x_n)\in V$ for $n>N$. This is just the definition of $f(x_n)\to f(x)$.

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Let $X$ be metrizable with metric $d$ and suppose $V$ is an open set of $Y$ with $f^{-1}(V)$ not open, that is, there exists $x\in f^{-1}(V)$ (thus $f(x)\in V$) such that $B(x,\frac1n))\not\subseteq f^{-1}(V)$ for all $n\in\mathbb{N}$. For each $n\in\mathbb{N}$ consider $x_n\in B(x,\frac1n)\setminus f^{-1}(V)$. One can prove that $x_n\to x$. If $f(x_n)\to f(x)$, there exists $N\in\mathbb{N}$ such that $f(x_n)\in V$ for $n>N$, which implies $f^{-1}f(x_n)\subseteq f^{-1}(V)$ for $n>N$, a contradiction since $x_n\in f^{-1}f(x_n)$ and $x_n\notin f^{-1}(V)$.

Answer 2

Hint: Take open $V$ and assume $f^{-1}(V)$ is not open. That means that there exists $x\in f^{-1}(V)$ such that for all $n$, $B_n:=B(x,1/n)\not\subseteq f^{-1}(V)$, i.e. for all $n$, there exists $x_n\in B_n$ and $x_n\not\in f^{-1}(V)$. Yet, $x_n\to x$. Arrive at contradiction from here.

Answer 3

Use the sequence lemma (as Munkres formulates it):

If $A \subseteq X$ and $x_n$ is a sequence from $A$ with $x_n \to x$, then $x \in \overline{A}$ and the reverse holds in metrisable spaces.

Then suppose $f: X \to Y$ is sequentially continuous and $X$ is metrisable then $f$ is continuous. We can also show continuity by showing $f^{-1}[C]$ to be closed for $C \subseteq Y$ closed. So let $C$ be closed and let $x \in \overline{f^{-1}[C]}$. By the sequence lemma (reverse) there is a sequence $x_n$ from $f^{-1}[C]$ that converges to $x$. So $f(x_n) \to f(x)$ by sequential continuity. Also, all $f(x_n) \in C$. As $C$ is closed by the sequence lemma again $f(x) \in \overline{C}=C$, so $x \in f^{-1}[C]$. As $x \in \overline{f^{-1}[C]}$ was arbitrary, $f^{-1}[C]$ is closed. Hence $f$ is continuous.