$X$ is a subset of $\mathbb{R}$. Prove that $|x-2|<c$ for all $x$ that are elements of $X$.

Question

Suppose that $X\subset\mathbb{R}$ (real numbers) is a bounded set. Prove that there exists a number $c$ that is an element of the real numbers, such that for all $x$ that are elements of $X$, the absolute value of $x-2$ is less than or equal to $c$.

can we assume $X$ is non-empty? OR

if $X$ is non empty, the $X$ is a subset of $[m,M]$ and every $m\le x \le M$ if $x$ is an element if X, and then let $c=$ max of the absolute value of either $M-2$ or $m-2$. and then show the absolute value of $x-2$ is between $-c$ and $c$?

Answer 1

Suppose $m \le x \le M$. Then $m-2 \le x - 2 \le M-2$.

If $x-2 \ge 0$, then $x-2 \le M-2$.

If $x-2 \le 0$, then $2-x \le 2-m$.

So either way $ |x-2| \le \max\{M-2, 2-m\} < \max\{M-2, 2-m\} + 1$.