Let $f:X\rightarrow\textbf{R}$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.

Question

Let $X$ be a subset of $\textbf{R}$, and let $f:X\rightarrow\textbf{R}$ be a uniformly continuous function. Suppose that $E$ is a bounded subset of $X$. Then $f(E)$ is also bounded.

MY ATTEMPT

I will post it later after I reformulate my answer.

Answer 1

Part of what you have written is false. For instance, you assume that $0 \in X$ and you do not know this! However, you have the right definitions in place we should work with these. Also, your last statement when you compare $\delta$ and $M$ seems like a strange approach. Remember that $\delta$ is likely really small and $M$ really large. While your line of reasoning is correct, in practice it is unlikely that $\delta \geq M$. It would imply that the function $f$ differs by $1$ everywhere on $E$...certainly a restrictive condition!

Nevertheless, you have the necessary components. Here is one possible answer that goes along your line of thinking: For simplicity, I will prove this in the case that $X$ is nonnegative.

Since $E$ is bounded, there is $M$ large enough so that $E \subset [0,M]$. By uniform continuity, there is $\delta > 0$ such that $x,y \in X$ with $|x-y| \leq \delta$ implies $|f(x) - f(y)| < 1$. Let $x_0 \in E$. Since $\delta > 0$ we find $m$ large enough so that

$$ E \subset [0,M] \subset \bigcup_{k=0}^m [k\delta,(k+1)\delta] $$

Think about why I wrote the union like this. Note each of the subintervals of this union has length less than $\delta$. Thus, for each $x,y$ in each of the subintervals the function $f$ can vary by at most $1$. For each $k$ such that $E \cap [k\delta,(k+1)\delta] \neq \emptyset$ define

$$ M_k = \sup_{x \in [k\delta,(k+1)\delta]} |f(x)| $$

Then, each $M_k < +\infty$ since $f$ can vary by at most $1$ on each subinterval (see if you can make this more rigorous!). Define $M_k = 0$ if $[k\delta,(k+1)\delta] \cap E = \emptyset$. Then, letting $M^* = \max(M_1,M_2,\ldots,M_m)$ we see that $|f(x)| \leq M^*$ on $\bigcup_{k=0}^m [k\delta,(k+1)\delta] \supset E$.