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!

  • $\begingroup$ How do you get from $x' > x$ to $X$ been unbounded? $\endgroup$ – mihaild May 26 at 20:11
  • $\begingroup$ @mihaild by picking an element of $x \in X$ I was able to get a higher element $x' \in X$. $\endgroup$ – Bruno Reis May 26 at 20:15
  • $\begingroup$ $x$ isn't arbitrary element from $X$. Also, for any element from $(0, 1)$ there is greater element from $(0, 1)$, but $(0, 1)$ is bounded. $\endgroup$ – mihaild May 26 at 20:16
  • $\begingroup$ In my proof, we can pick an arbitrary $\epsilon_0, \epsilon' > 0$. That implies that $x' - x > 0$ can be arbitrarily large, hence $X$ is unbounded above. $\endgroup$ – Bruno Reis May 26 at 20:45
  • $\begingroup$ We can't pick them arbitrary - we get them from non boundness of $\overline{X}$. We can pick them arbitrary large, of course (we actually need only $\epsilon_0$ to be large), but I think for the proof to be valid it should be said. And any my attempt to say it formally leads (after removing unnecessary variables) to something like "there is arbitrarily large $a \in \overline X$ and $x \in X$ close to $a$, so there are arbitrarily large numbers in $X$". $\endgroup$ – mihaild May 26 at 20:58

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,$$\left(\forall x\in\overline X\right):x\leqslant M.$$

  • $\begingroup$ Simple, direct and elegant. Thanks! $\endgroup$ – Bruno Reis May 26 at 20:12
  • $\begingroup$ Sometimes the contrapositive is overly seductive. $\endgroup$ – Lee Mosher May 26 at 20:43
  • $\begingroup$ @LeeMosher Yeah, you're right. But for you, what I've done is correct? $\endgroup$ – Bruno Reis May 26 at 20:46

$X \subset \mathbb{R}$ 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.


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