Continuity of a Function in Terms of Closure and Interior I've managed to show that the following are equivalent (where $f^*$ and $f_*$ are the preimage and image of f respectively):
$\bullet$ $f:X \rightarrow Y$ is continuous
$\bullet$ $f^*(S^{\circ}) \subseteq f^*(S)^{\circ}$
$\bullet$ $\overline{f^*(S)} \subseteq f^*(\overline{S})$
$\bullet$ $f_*(\overline{S}) \subseteq \overline{f_*(S)}$
Is the following also equivalent: $f_*(S)^\circ \subseteq f_*(S^\circ)$ (or something similar)? I've not managed to get very far, although my proof of condition 3 hinged on the fact that preimages, complements, interiors and closures all behave well together, whereas this is not case for the image of a function, so I feel this may not be true.
 A: tkr provided an example that shows that int$(f(S))\subseteq f($int$(S)$ does not imply continuity. To see that continuity does not imply the property consider $f:\Bbb R\to\Bbb R_{\ge0}$,
$f(x)=\begin{cases}
0, \text{ if }x\le0\\
x, \text{ if }x\ge0
\end{cases}$
Let $S=[0,1]$. Then int$(f(S))=[0,1)$, but $f($int$(S))=(0,1)$
On the other hand, this map $f$ considered as a map $\Bbb R\to\Bbb R$ does not satisfy $f($int$(S))\subseteq$ int$(f(S))$ (the reverse inclusion), since for $S=[-1,1]$ the former is $[0,1)$ while the later is $(0,1)$.
To see that this inclusion does not imply continuity, consider the map $g:\Bbb R\to \{0,1\}$
$g(x)=\begin{cases}
0 \text{, if }x<0\\
1 \text{, if }x\ge0
\end{cases}$
But you can show that the later inclusion is equivalent to openness of $f$. The former inclusion is rather useless.
A: The continuous image of a non-empty set with empty interior may have non-empty interior, so we may have $(f_*(S))^o\not=\phi$ and $f_*(S^o)=\phi$. For example, if f is the Devil's Staircase function, which maps [0,1] onto itself continuously,and S is the Cantor set,then f(S)=[0,1] and int(S)=$\phi$.
