Counter example of continous function such that there is Set S with $f(S)^\circ \subset (f(S^\circ))$

Question

$S^\circ$ denotes the interior of a set $S$. Is there an example of a continuous function $f$ and a set S with $(f(X))^\circ \not\subset f(S^\circ)$ ?

I know that$f(S^\circ)\subset (f(S))^\circ$ is not always true; for example

$$f(x)=x, ....[0,1]$$

$$f(x)=x-1 ....[2,3]$$

I tried hard but I could not find counterexample for $(f(S))^\circ\subset f(S^\circ)$

Any help will be appreciated

Answer 1

This may be overkill, but the cantor function with $S=$ the middle-thirds cantor set will work:

2 properties of this function are: $S^\circ = \emptyset$ and $f(S)=[0,1]$