# [SOLVED] Prove: if f(A) open subset of R the R is continuous

• Feb 10th 2009, 02:56 AM
Amanda1990
[SOLVED] Prove: if f(A) open subset of R the R is continuous
Let A be an open subset of $\Re$ and let $f: A \rightarrow \Re$ be an increasing function. Prove that if f(A) is an open subset of $\Re$ then f is continuous.
• Feb 10th 2009, 09:42 AM
ThePerfectHacker
Quote:

Originally Posted by Amanda1990
Let A be an open subset of $\Re$ and let $f: A \rightarrow \Re$ be an increasing function. Prove that if f(A) is an open subset of $\Re$ then f is continuous.

Set $B=f(A)$, we are told that $A\text{ and }B$ are open subsets of $\mathbb{R}$.

Let $a\in A$, so $f(a) \in B$ but $B$ is open so there exists $r>0$ so that $(f(a)-r,f(a)+r) \subseteq B$. From here and on we will prove that the $\delta,\epsilon$ definition of continuity is satisfied for all $\epsilon < r$. Since $f(a) - \epsilon,f(a)+\epsilon \in B$ it means there exists $a_1\in A \text{ and }a_2\in A$ so that $f(a_1) = f(a) - \epsilon \text{ and }f(a_2) = f(a) + \epsilon$. Now if $a_1 < x < a_2 \implies f(a_1) < f(x) < f(a_2)$ and so $f(a) - \epsilon < f(x) < f(a) + \epsilon \implies |f(x) - f(a)| < \epsilon$. Certainly, $a_1 < a < a_2$, so the neighborhood $N = \{ x\in \mathbb{R} | a_1 < x < a_2 \}$ of $a$ gets mapped within $\epsilon$ of $f(a)$. Thus, $f$ is continous at any point $a\in A$.