# Metric Spaces - Open subset

1. Feb 5, 2012

### Jonmundsson

1. The problem statement, all variables and given/known data
Let $a,b \in \mathbb{R}$ and we define $I = [a,b]$ ([] means closed set). Let $\mathcal{C}_{\mathbb{K}}(I)$ be the space of all continuous functions $I \to \mathbb{K}$ with the norm $f \mapsto ||f||_I = \displaystyle \sup _{x \in I} |f(x)|$. Let $U$ be the set of all continuous functions $I \to \mathbb{K}$ so that $\displaystyle |\int _a ^b f(x)dx| < 1$. Show that $U$ is an open subset of $\mathcal{C}_{\mathbb{K}}(I)$.

2. Relevant equations
If $f: X \to Y$ is a continuous function and $B \subset Y$ is open then the preimage $f^{-1}$ is open.

For $B \subset Y$ the set $f^{-1} = \{x \in X : f(x) \in B\}$ is called the preimage of B.

$\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$

3. The attempt at a solution

The function $\mathcal{C} _{\mathbb{K}} (I) \to \mathbb{K}, f \mapsto \displaystyle \int _a ^b f(x)dx$ is continuous because

$\displaystyle |\int _a ^b f(x)dx| \leq \int _a ^b |f(x)|dx \leq \int _a ^b ||f||_I dx = ||f||_I \int _a ^b dx = (b-a)||f||_I$

I want to show that preimage of U is open then U is open using the two definitions I listed above. I'm just curious whether that is a legit method to solve this problem. I'm also wondering what the purpose of the norm is in this problem. If anyone can answer these questions I will be grateful.

Thanks.