Let $A = \{x : x \in \mathbb{Q^{+}}, \space x^2<2 \}$, Show that $\sup (A)= \sqrt 2$

Question

I am trying to get ready to go back to school for mathematics after a long break and I find the solution in my book to be unfathomably complicated for this for no reason. I am wondering if my solution makes sense and if so why it isn't how they decided to do it.

Firstly, since $x^2 <2$ it is clear that $x < \sqrt 2 \space$, this tells us that $\sqrt 2 $ is an upper bound for A. Lets assume that $\sqrt 2 $ is not the least upper bound, then there exists some $a< \sqrt2 \space$ that is the least upper bound of A. Let us choose a $c \in (a,\sqrt 2 )\space $ where $c$ is rational (It is clear that there are infinitely many rationals between these two numbers.)

Since $c$ is rational and $c < \sqrt 2 $ we know that $ c \in A $ this is a contradiction. So it must be the case that $\sqrt 2 $ is the least upper bound of A.

The books solution seems to be to create a sequence that is in A and converges to $\sqrt 2$ but its very complicated to follow and quiet long. Does that mean that my solution is incorrect?

it would depend on what you already know. What is the definition of $\sqrt{\bullet}$? is it as a function from $[0,\infty)\to \mathbb R$? You should check the book to see if everything that you used has been proven starting from the base assumptions (and for the same reason, you should say what book you're using. Rudin?) — Calvin Khor, Sep 04 '21 at 05:01
It is indeed possible to construct a sequence in $\mathbb{Q}$ that is $<\sqrt{2}$ and has $\sqrt{2}$ as a limit. — rtybase, Sep 04 '21 at 08:02

score 1 · Accepted Answer · answered Sep 04 '21 at 04:56

Your solution is correct. Let me make an attempt to explain the sequential solution though. We have $A = \{x : x \in \mathbb{Q^{+}}, \space x^2<2 \}$, and as you noticed, for every $x\in A$, $x < \sqrt 2$. It follows that $\sup A \le \sqrt 2$. The job of the "sequence approach" is to now show that $\sup A \ge \sqrt 2$, in order to conclude $\sup A = \sqrt 2$.

Since the rationals are dense in reals, we can find $a_n\in \mathbb Q^+$ such that $\sqrt 2 - \frac1n < a_n < \sqrt2$. Clearly, $a_n \in A$ for all $n\in \mathbb N$. By squeeze theorem, it follows that $\lim_{n\to\infty} a_n = \sqrt 2$. Now, the supremum of $A$ is certainly at least as large as every element of $A$, i.e. $\sup A \ge a_n$ for all $n\in \mathbb N$. Taking limits as $n\to\infty$, we obtain $\sup A \ge \lim_{n\to\infty} a_n = \sqrt2$.

As stated earlier, $\sup A \ge \sqrt 2$ and $\sup A \le \sqrt 2$ imply $\sup A = \sqrt 2$.

This is a much better solution than what my problem book has, thanks for the insight. — Faust, Sep 04 '21 at 17:56

Let $A = \{x : x \in \mathbb{Q^{+}}, \space x^2<2 \}$, Show that $\sup (A)= \sqrt 2$

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