06 - Images of a Continuous Functions is Bounded

Images of a Continuous Functions is Bounded

Let $f:U\to {\mathbb{R}}^2\to \mathbb{R}$ be a continuous function
where $U$ is non-empty and compact then

$$f(U)\text{ is bounded and achieves that boud}$$

That is $\exists b\in \mathbb{R}$ and $a\in U$ such that

$$|f(x)|\le b\text{ for all }x\in U\text{ and }|f(a)|=b$$

Proof - Uses Lecture Topic 10 $U$ is compact then it is sequentially compact

Since

As $|f|$ is continuous (composition of two continuous functions)
Then $|f(U)|$ is non-empty and bounded so it has a supremum
Since $|f(U)|$ is closed then the supremum is in $|f(U)|$
Hence there exists $b\in \mathbb{R}$ and $a\in U$ such that $|f(x)|\le b$ for all $x\in U$ and $|f(a)|=b$