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

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

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

2. Originally Posted by Amanda1990
Let A be an open subset of $\displaystyle \Re$ and let $\displaystyle f: A \rightarrow \Re$ be an increasing function. Prove that if f(A) is an open subset of $\displaystyle \Re$ then f is continuous.
Set $\displaystyle B=f(A)$, we are told that $\displaystyle A\text{ and }B$ are open subsets of $\displaystyle \mathbb{R}$.

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