Given are two metric spaces $(X , d_X)$ , $(Y , d_Y)$ and a continuous function $f : X \rightarrow Y$.
For all subsets $A \subseteq X $, is $f(A)^{°} \subseteq f(A^{°})$ true? ( ' $^{°}$ ' means interior)
My first thought was that $f(A^{°})$ is open because $A^{°}$ is. By definition of interior $f(A)^{°}$ is the largest open subset of $f(A)$. So $f(A)^{°}$ can never be a subset of $f(A^{°})$, is that right?
I tried finding a counterexample but couldn't find an obvious one that came to mind. Anyone has an idea? Thanks in advance!
Think of a function $f$ that takes the line $(-\infty,\infty)$ onto the unit circle $S^1$, where $f$ is one-to-one on an $[0,1]$ except at the endpoints, where $f(0) = f(1)$. For instance $$f(t) = (\cos 2\pi t,\sin 2\pi t)$$ will do the job.
Let $A = [0,1]$. Since $f(A) = S^1$ you also have $f(A)^\circ = S^1$.
On the other hand, $f(A^\circ)$ is missing a point of $S^1$.