What is the most direct proof of $f$ is continuous iff $f\left(\overline{A}\right) \subset \overline{f(A)}$? Here is our definition of continuity:

Let $X$ and $Y$ be any topological spaces, let $f \colon X \rightarrow Y$ be a mapping, and let $p$ be a point of $X$. Then $f$ is said to be continuous at point p if, for every open set $V$ of $Y$ such that $f(p) \in V$, there exists an open set $U$ of $X$ such that $p \in U$ and $f(U) \subset V$.


Let $S$ be any subset of $X$. If $f$ is continuous at every point of $S$, then $f$ is said to be continuous on set $S$.


And, if $f$ is continuous at every point of $X$, then $f$ is said simply to be continuous.

Then what is the most direct way of proving the following statement?

Let $X$ and $Y$ be any topological spaces. Then a mapping $f \colon X \rightarrow Y$ is continuous (at every point of $X$) if and only if, for every subset $A$ of $X$,
$$
f \left( \overline{A} \right) \subset \overline{f(A)},
$$
where on the left-hand-side we have the colsure of $A$ in the topological space $X$ and on the right-hand-side we have the closure of $f(A)$ in the topological space  $Y$.

My Attempt:

Suppose that $f \colon X \rightarrow Y$ is continuous. Let $q$ be any point of $f\left(\overline{A}\right)$. We show that this point $q \in \overline{f(A)}$.


Let $V$ be any open set of $Y$ such that $q \in V$.  In order to show that  $q \in \overline{f(A)}$, we need to show that $V \cap f(A) \neq \emptyset$.


Now as $q \in f\left( \overline{A} \right)$, so there exists a point $p \in \overline{A}$ such that $q = f(p)$; moreover as $p \in \overline{A}$ and $\overline{A} \subset X$, so $p \in X$; and as $p \in X$ and $f$ is continuous at every point of $X$, so $f$ is continuous at $p$ also.


Thus the mapping $f \colon X \rightarrow Y$ is continuous at point $p \in X$ and $V$ is an open set of $Y$ containing $f(p)$. So there exists an open set $U$ of $X$ such that $p \in U$ and $f(U) \subset V$.


Now as $p \in \overline{A}$ and $U$ is an open set of $X$ containing $p$, so we must have $U \cap A \neq \emptyset$; let $a \in U \cap A$.


Then $a \in A$ and $a \in U$, which implies that $f(a) \in f(A)$ and $f(a) \in f(U)$, but $f(U) \subset V$, so we can conclude that $f(a) \in V$ also. Thus we have $f(a) \in f(A) \cap V$, which implies that $f(A) \cap V \neq \emptyset$.


So far we have shown that, for every open set $V$ of the topological space $Y$ such that $q \in V$, we have $f(A) \cap V \neq \emptyset$. Therefore $q \in \overline{f(A)}$. But $q$ was an arbitrary point of set $f\left(\overline{A}\right)$. Hence we can conclude that
$$
f \left( \overline{A} \right) \subset \overline{f(A)}. 
$$

Am I right?

Conversely, suppose that, for every subset $A$ of $X$, we have
$$
f \left( \overline{A} \right) \subset \overline{f(A)}. 
$$
We show that $f$ is continuous (at every point of $X$). Let $p$ be an arbitrary point of $X$. We show that $f$ is continuous at $p$. For this, let $V$ be any open set of $Y$ such that $f(p) \in V$. Then $Y\setminus V$ is a closed set of $Y$ and $f(p) \not\in Y \setminus V$.


As $Y \setminus V$ is a closed set of $Y$, so
$$ \overline{Y \setminus V} = Y \setminus V, $$
which implies that
$$
f^{-1} \left( \overline{Y \setminus V}  \right) = f^{-1} (Y \setminus V) = f^{-1}(Y) \setminus f^{-1}(V) = X \setminus f^{-1} (V). 
$$

Is my work up to this point correct? If so, then how to proceed from here? Or, are there any mistakes in what I have done?
 A: Let $f:X\to Y$ be continuous and let $A\subseteq X$.
Then $f^{-1}\left(\overline{f\left(A\right)}\right)$
is closed since it is the preimage of a closed set.
This evidently
with $A\subseteq f^{-1}\left(\overline{f\left(A\right)}\right)$ so
we are allowed to conclude that $\overline{A}\subseteq f^{-1}\left(\overline{f\left(A\right)}\right)$
or equivalently $f\left(\overline{A}\right)\subseteq\overline{f\left(A\right)}$.

Let $f:X\to Y$ be not continuous.
Then some closed set $B\subseteq Y$ exists such that
$A:=f^{-1}\left(B\right)$ is not closed.
Then $\overline{A}-A$ will
contain an element $x$.
Then $f\left(x\right)\notin B$ because $x\notin A=f^{-1}(B)$.
Observe that $f\left(A\right)\subseteq B$ so that - because $B$ is closed - we have: $\overline{f\left(A\right)}\subseteq B$.
We conclude that $f\left(x\right)\notin\overline{f\left(A\right)}$.
But $x\in\overline A$ so that $f(x)\in f(\overline A)$ so this shows that we do not have $f\left(\overline{A}\right)\subseteq\overline{f\left(A\right)}$.
So it has been proved that whenever $f$ is not continuous we can find a set $A$ such that $f\left(\overline{A}\right)\subseteq\overline{f\left(A\right)}$ is not true.
A: Let $X$ and $Y$ be topological spaces and $f \colon X → Y$ be map. We say

*

*$f$ is drag-continuous if for $V ⊆ Y$ open $f^{-1}(V)$ is open in $X$.

*$f$ is touch-continuous if for $T ⊆ X$ arbitrary $f(\overline T) ⊆ \overline {f(T)}$.

Drag-continuous maps are ones so that for any open $V ⊆ Y$ and any $x ∈ X$ with $f(x) ∈ V$, they drag an entire neighbourhood $U ⊆ X$ of $x$ into $V$, that is $f(U) ⊆ V$. Touch-continuous maps are ones so that if $x ∈ X$ touches a part $T ⊆ X$, that is $x ∈ \overline T$, then $f(x)$ touches $f(T)$, that is $f(x) ∈ \overline {f(T)}$.
To prove they are equivalent, remember the basic facts that

*

*for arbitrary maps $f \colon X → Y$ and $A ⊆ X$ and $B ⊆ Y$, we have
$f(A) ⊆ B \iff A ⊆ f^{-1} (B)$,

*being drag-continuous is equivalent to preimages of closed sets being closed,

*for sets $A ⊆ X$ and $T ⊆ X$ closed, $A ⊆ T \iff \overline A ⊆ T$, and

*for sets $T ⊆ X$, we have $T ⊆ X~\text{is closed} \iff \overline T ⊆ T$.


Let $f$ be drag-continuous and $A ⊆ X$. Then
$$f(\overline A) ⊆ \overline {f(A)} \iff \overline A ⊆f^{-1} (\overline {f(A)}) \iff A ⊆ f^{-1}(\overline {f(A)}) \iff f(A) ⊆ \overline {f(A)},$$
so indeed $f(\overline A) ⊆ \overline {f(A)}$, as the latter inclusion is true by extensivity of closing.
Let $f$ be touch-continuous and $B ⊆ X$ closed. Then
$$f^{-1}(B) ⊆ X ~\text{is closed} \iff \overline{f^{-1}(B)} ⊆ f^{-1} (B) \iff f(\overline{f^{-1}(B)}) ⊆ B.$$
Now for $A = f^{-1}(B)$ we have $f(A) ⊆ B$, and since $B$ is closed, $\overline {f(A)} ⊆ B$, so
$$f(\overline A) ⊆ \overline{f(A)} ⊆ B, \quad\text{so}\quad f(\overline{f^{-1}(B)}) ⊆ B,$$
hence $f^{-1}(B) ⊆ X$ is indeed closed.
A: Suppose that $f$ is continuous, and let $A \subseteq X$ be any subset.
$\overline{f[A]}$ is closed in $Y$ and contains $f[A]$ and so by continuity, $f^{-1}[\overline{f[A]}]$ is closed and it clearly contains $A$.
So $\overline{A} \subseteq  f^{-1}[\overline{f[A]}]$ (the closure is the smallest closed superset of $A$) and so $f[\overline{A}] \subseteq \overline{f[A]}$  by definition.
OTOH if $f$ fulfills the closure condition, let $C \subseteq Y$ be closed.
Define $A= f^{-1}[C]$ and by the property, $$f[\overline{A}] \subseteq \overline{f[A]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C}=C$$
as $C$ is closed and this implies that $\overline{A} \subseteq f^{-1}[C]=A$ and thus $A$ is closed and $f$ is continuous (inverse image of a closed set is closed).
