# need help

• Apr 17th 2009, 10:07 PM
mancillaj3
need help
consider a function $\displaystyle f: Real \rightarrow Real$. Prove that f is continuous on R iff for every subset B of Real
$\displaystyle f^{-1}(int B) \subseteq f^{-1} (B)$
• Apr 18th 2009, 01:22 AM
clic-clac
Hi

This is always true.

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

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

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

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

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

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