1. ## need help

consider a function $f: Real \rightarrow Real$. Prove that f is continuous on R iff for every subset B of Real
$f^{-1}(int B) \subseteq f^{-1} (B)$

2. Hi

This is always true.

I think you meant: $f^{-1}(intB)\subseteq int(f^{-1}(B))$

I $f$ is continuous, then for any open set $V,\ f^{-1}(V)=U$ is an open set. So let $x$ be an element of $f^{-1}(intB),$ i.e. there exists an open set $V$ such that $f(x)\in V\subseteq B.$ Then $f^{-1}(V)$ is an open subset of $f^{-1}(B)$ and contains $x,$ therefore $x\in int(f^{-1}(B))$ and we can conclude: $f$ continuous $\Rightarrow f^{-1}(intB)\subseteq int(f^{-1}(B))$

Conversely, assume f $^{-1}(intB)\subseteq int(f^{-1}(B))$ . Let $U$ be an open set. Then intU=U, and we can write:

$f^{-1}(U)=f^{-1}(intU)\subseteq int(f^{-1}(U))$ i.e. $f^{-1}(U) \subseteq int(f^{-1}(U))$

But, for any subset $A,\ A\subseteq intA$ means that $A$ is open (if you never saw that, try to prove it)

Thus for any open subset $U$, $f^{-1}(U)$ is open so f is continuous, we just proved: $f^{-1}(intB)\subseteq int(f^{-1}(B)) \Rightarrow$ $f$ continuous.