$f$ continuous$⇔f(\overline A)\subseteq\overline{f(A)}\;∀A⊆X$

hbghlyj

Wilson A. Sutherland - Introduction to Metric and Topological Spaces-Oxford University Press (2009)
Proposition 6.12 A map $f:X\to Y$ of metric spaces is continuous iff $f(\overline A)\subseteq\overline{f(A)}$ for all subsets $A\subseteq X$.
Proof.
First suppose that $f: X \rightarrow Y$ is continuous, and let $y \in f(\overline{A})$ for some $A \subseteq X$ and $\varepsilon>0$. Then $y=f(x)$ for at least one $x \in \overline{A}$. By continuity of $f$ at $x$, there exists $\delta>0$ such that $f\left(B_\delta(x)\right) \subseteq B_{\varepsilon}(y)$. By definition of $\overline{A}$ there exists some $a \in A \cap B_\delta(x)$. Then $f(a) \in f\left(B_\delta(x)\right) \subseteq B_{\varepsilon}(y)$. So $f(a) \in B_{\varepsilon}(y) \cap f(A)$ which shows that $y \in \overline{f(A)}$. Hence $f(\overline{A}) \subseteq \overline{f(A)}$

Conversely suppose that $f(\overline{A}) \subseteq \overline{f(A)}$ for any subset $A$ of $X$. We shall prove that the inverse image of any closed subset $V \subseteq Y$ is closed in $X$, so that $f$ is continuous by Proposition 6.6. For $f\left(\overline{f^{-1}(V)}\right) \subseteq \overline{f\left(f^{-1}(V)\right)}$, and $f\left(f^{-1}(V)\right) \subseteq V$ so $\overline{f\left(f^{-1}(V)\right)} \subseteq \overline{V}=V$, where the equality follows from Proposition 6.11(c) since $V$ is closed in $Y$. Hence $f\left(\overline{f^{-1}(V)}\right) \subseteq V$, so $\overline{f^{-1}(V)} \subseteq f^{-1}(V)$. Since we always have $f^{-1}(V) \subseteq \overline{f^{-1}(V)}$, this shows that $\overline{f^{-1}(V)}=f^{-1}(V)$, and $f^{-1}(V)$ is closed in $X$ by Proposition 6.11(c).

hbghlyj

Exercise 6.20 A map $f:X\to Y$ of metric spaces is continuous iff $f^{-1}(\operatorname{Int}B)\subseteq\operatorname{Int}f^{-1}(B)$ for all subsets $B\subseteq Y$.
Proof.
First suppose that $f: X \rightarrow Y$ is continuous, and let $B \subseteq Y$. By Proposition 6.21(e) the interior $\operatorname{Int}B$ of $B$ is open in $Y$, so by continuity, $f^{-1}(\operatorname{Int}B)$ is open in $X$. Now $\operatorname{Int}B \subseteq B$, so $f^{-1}(\operatorname{Int}B) \subseteq f^{-1}(B)$. This gives that $f^{-1}(\operatorname{Int}B)$ is open in $X$ and contained in $f^{-1}(B)$, hence by Proposition 6.21(f) it is contained in the interior of $f^{-1}(B)$ as required.

Conversely, suppose that $f^{-1}\left(\operatorname{Int}B\right)$ is contained in the interior of $f^{-1}(B)$ for any subset $B$ of $Y$. In particular take $B$ to be open in $Y$. By Proposition 6.21(c) then $\operatorname{Int}B=B$, so $f^{-1}(\operatorname{Int}B)=f^{-1}(B)$. So by the given condition, $f^{-1}(B)$ is contained in the interior of $f^{-1}(B)$. But the interior of any set $A \subseteq X$ is contained in $A$, so $f^{-1}(B)$ equals its interior, which by Proposition 6.21(c) shows that $f^{-1}(B)$ is open in $X$. This shows that $f$ is continuous.

hbghlyj

原书是使用 $\mathring A$ 表示"interior of A", 我在上面都给替换成 Int A 了.
用 \mathring A 的好处是看上去跟 $\bar A$ 比较像(都是在上面加的符号), 容易联想到它们是对偶的:$$\overline{X\setminus A}=X\setminus\mathring A$$
用 Int A 的好处是可以给一个很长的式子加括号再加 Int 比如$\operatorname{Int}(A∩B)$,
但写成 \mathring{A∩B} 即 $\mathring{A∩B}$ 就不知道 \mathring 的作用范围了 (看上去还以为是A \mathring ∩ B)
因为圆圈不像\overline可以拉长

hbghlyj

Topology notes
Proposition 1.36. A map $f: X \rightarrow Y$ of topological spaces is continuous if and only if $f(\bar{A}) \subseteq \overline{f(A)}$ for every $A \subseteq X$.

Proof. Assume that $f$ is continuous. Since $\overline{f(A)}$ is closed, it follows that $f^{-1}(\overline{f(A)})$ is closed. The latter subset also contains $A$, therefore it contains $\bar{A}$. We have thus proved that $\bar{A} \subseteq f^{-1}(\overline{f(A)})$, which implies that $f(\bar{A}) \subseteq \overline{f(A)}$.
Conversely, assume that $f(\bar{A}) \subseteq \overline{f(A)}$ for every $A \subseteq X$.
Let $F$ be a closed subset in $Y$ and let $A=f^{-1}(F)$. Then $f(A) \subseteq F$, whence $f(\bar{A}) \subseteq \overline{f(A)} \subseteq F$. It follows that $\bar{A} \subseteq f^{-1}(F)=A$, therefore $\bar{A}=A$. This implies that $A$ is closed.

hbghlyj

inclusion can be strict:
$f(x)=x^{-1},A=[1,+\infty),f(A)=(0,1]$
$f(\overline{A})=(0,1]\subsetneq \overline{f(A)}=[0,1]$