# testing for continuity via closure operation

###### Proposition 1.

Let $X,Y$ be topological spaces, and $f:X\to Y$ a function. Then the following are equivalent:

1. 1.

$f$ is continuous,

2. 2.

for any closed set $D\subseteq Y$, the set $f^{-1}(D)$ is closed in $X$,

3. 3.

$f(\overline{A})\subseteq\overline{f(A)}$, where $\overline{A}$ is the closure of $A$,

4. 4.

$\overline{f^{-1}(B)}\subseteq f^{-1}(\overline{B})$,

5. 5.

$f^{-1}(C^{\circ})\subseteq f^{-1}(C)^{\circ}$, where $C^{\circ}$ is the interior of $C$.

###### Proof.
• $(1)\Leftrightarrow(2)$. Use the identity $f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$ for any function $f$. Then $f^{-1}(Y-D)=X-f^{-1}(D)$. So if $D$ is closed (or open), $f^{-1}(Y-D)$ is open (or closed), whence $f^{-1}(D)$ is closed (or open).

• $(2)\Leftrightarrow(3)$. Suppose first that $f:X\to Y$ is continuous. Since

 $\overline{f(A)}=\bigcap\{C\mid C\mbox{ closed in }Y,\mbox{ and }f(A)\subseteq C\},$

$A\subseteq f^{-1}f(A)\subseteq f^{-1}(C)$, which is closed in $X$. So $\overline{A}\subseteq f^{-1}(C)$, and therefore $f(\overline{A})\subseteq ff^{-1}(C)\subseteq C$. As a result,

 $f(\overline{A})\subseteq\bigcap\{C\mid C\mbox{ closed in }Y,\mbox{ and }f(A)% \subseteq C\}=\overline{f(A)}.$

Conversely, let $V$ be closed in $Y$. Then $\overline{V}=V$. Let $U=f^{-1}(V)$. So $f(U)=V$. Let $W=\overline{U}$. Then $f(W)=f(\overline{U})\subseteq\overline{f(U)}=\overline{V}=V$. So $W\subseteq f^{-1}f(W)\subseteq f^{-1}(V)=U\subseteq\overline{U}=W$. As a result, $U=W$ is closed.

• $(3)\Leftrightarrow(4)$. First, assume $(2)$. Let $B\subseteq Y$ and $A=f^{-1}(B)$. So $f(A)\subseteq B$. Then $f(\overline{A})\subseteq\overline{f(A)}\subseteq\overline{B}$. As a result, $\overline{f^{-1}(B)}=\overline{A}\subseteq f^{-1}f(\overline{A})\subseteq f^{-% 1}(\overline{B})$.

Conversely, assume $(3)$. Let $A\subseteq X$ and $B=f(A)$. So $A\subseteq f^{-1}(B)$. Then

 $f(\overline{A})\subseteq f(\overline{f^{-1}(B)})\subseteq ff^{-1}(\overline{B}% )\subseteq\overline{B}=\overline{f(A)}.$
• $(4)\Leftrightarrow(5)$. First, assume $(3)$. We use the identity: $C^{\circ}=Y-\overline{Y-C}$. Then

 $\displaystyle f^{-1}(C^{\circ})$ $\displaystyle=$ $\displaystyle f^{-1}(Y-\overline{Y-C})=f^{-1}(Y)-f^{-1}(\overline{Y-C})% \subseteq X-\overline{f^{-1}(Y-C)}$ $\displaystyle=$ $\displaystyle X-\overline{f^{-1}(Y)-f^{-1}(C)}=X-\overline{X-f^{-1}(C)}=f^{-1}% (C)^{\circ}.$

Conversely, assume $(4)$. We use the identity $\overline{B}=Y-(Y-B)^{\circ}$. Then

 $\displaystyle\overline{f^{-1}(B)}$ $\displaystyle=$ $\displaystyle X-(X-f^{-1}(B))^{\circ}=X-f^{-1}(Y-B)^{\circ}$ $\displaystyle\subseteq$ $\displaystyle X-f^{-1}((Y-B)^{\circ})=f^{-1}(Y-(Y-B)^{\circ})=f^{-1}(\overline% {B}).$

Title testing for continuity via closure operation TestingForContinuityViaClosureOperation 2013-03-22 19:09:11 2013-03-22 19:09:11 CWoo (3771) CWoo (3771) 8 CWoo (3771) Result msc 26A15 msc 54C05