Show that if $X \subset R$ is bounded above, then $\overline{X}$ is also bounded above.

Question

Show that if $X \subset R$ is bounded above, then $\overline{X}$ is also bounded above.

Here $\overline{X} = \{a\in \mathbb{R} \text{ | if }a\text{ is an adherence point of }X \}$

My Proof by contrapositive:

Suppose that $\overline{X}$ is not bounded above, therefore if $a \in \overline{X}$ then $\exists \epsilon > 0$ such that $(a+\epsilon) \in \overline{X}$.

Let $a \in \overline{X}$. Since $a$ is point of adherence of $X$, it follows that $\forall \epsilon > 0$ $X \cap (a-\epsilon, a+\epsilon) \neq \emptyset$. Let's call $x$ a value of $X$ in that open interval.

Since $\overline{X}$ is unbounded above, $\exists \epsilon_0 > 0$ such that $(a+\epsilon_0) \in \overline{X}$ which implies that $\forall \epsilon' > 0$ $X \cap \big( (a+\epsilon_0) - \epsilon', (a+\epsilon_0) + \epsilon' \big) \neq \emptyset$

Since $X \cap \big( (a+\epsilon_0) - \epsilon', (a+\epsilon_0) + \epsilon' \big) \neq \emptyset$, Let's call $x'$ a value of $X$ in that open interval.

Now let's choose $\epsilon < \epsilon_0 - \epsilon'$. Hence: $a + \epsilon < a + \epsilon_0 - \epsilon'$. With that we guarantee that every element of $X$ in the second interval is bigger than every element of $X$ in the first open interval.

Finally it follows that $x' > x$. Therefore $X$ is unbounded above.

---

I really think that my proof is valid and correct. What do you think? I'd appreciate any kind of constructive critics to help me to improve my approach. Thanks!

Answer 1

It looks correct, but it is much easier to prove it as follows: let $M$ be an upper bound of $X$. Then$$(\forall x\in X):x\leqslant M$$and therefore, since $(-\infty,M]$ is a closed set, $\overline X\subset(-\infty,M]$. In other words,$$\left(\forall x\in\overline X\right):x\leqslant M.$$

Answer 2

$X \subset \mathbb{R}$ is bounded above, i .e.

there is a $M$, real, s.t. for $x \in X $: $x <M;$

Let $y \in \overline{X}.$

There exists a sequence $x_n \in X$ s.t.

$\lim_{n \rightarrow \infty} x_n =y.$

Since $x_n < M$, $n \in \mathbb{N}$, we have

$y =\lim_{n \rightarrow \infty} x_n \le M$,

hence $\overline{X}$ is bounded.