| Version 5 |
Version 4 |
| Suppose $X$ is a topological space, and let $\overline{A}$ be the |
Suppose $X$ is a topological space, and let $\overline{A}$ be the |
| closure of $A$ in $X$. |
closure of $A$ in $X$. |
| Then the following properties hold: |
Then the following properties hold: |
|
|
| \begin{enumerate} |
\begin{enumerate} |
| \item $\overline{A}=A\cup A'$ where $A'$ is the derived set of $A$. |
\item $\overline{A}=A\cup A'$ where $A'$ is the derived set of $A$. |
| \item $A\subseteq \overline{A}$, and $A=\overline{A}$ if and only if $A$ |
\item $A\subseteq \overline{A}$, and $A=\overline{A}$ if and only if $A$ |
| is closed |
is closed |
| \item $\overline{A}=\emptyset$ if and only if $A=\emptyset$. |
\item $\overline{A}=\emptyset$ if and only if $A=\emptyset$. |
|
\item If $Y$ is another topological space, then $f\colon X \to Y$ is a continuous map,
|
\item If $Y$ is another topological space, and $f\colon X \to Y$ is a continuous map,
|
|
if and only if $f(\overline{A}) \subseteq \overline{f(A)}$. If $f$ is also a homeomorphism,
|
then $f(\overline{A}) \subseteq \overline{f(A)}$. If $f$ is also a homeomorphism,
|
| then $f(\overline{A}) = \overline{f(A)}$. |
then $f(\overline{A}) = \overline{f(A)}$. |
| \item If $E\subseteq X$ is any set, then |
\item If $E\subseteq X$ is any set, then |
| $$ |
$$ |
| A\cap \overline{E} \subseteq \overline{A\cap E}. |
A\cap \overline{E} \subseteq \overline{A\cap E}. |
| $$ |
$$ |
| \end{enumerate} |
\end{enumerate} |
