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I'm trying to work through this proof, and have hit a stumbling block.

This is what I have so far:

Let $0<k \in \mathbb{R}$, where $E = \{ x \in \mathbb{R} | x^{2} < k\}$.

First, we must show that $E$ is nonempty: Consider $0 \in \mathbb{R}$. Since $0^{2} = 0$ and $0 < 1$ (both proven previously), we have that $0 \in E$.

Next, we must show that $E$ is bounded above: Consider first the case where $k \geq 1$. Suppose $E$ is not bounded above - i.e., let $x \in E$, $x > k$. Then, $x^{2} > k^{2} > k$ (also due to a previously proven result). So, $x \notin E$, a contradiction. Therefore, $E$ is bounded above.

Consider next the case where $0 < k < 1$, then since $k < 1$, $E$ is clearly bounded above by $1$. (Also, not sure of this part. Seems a bit hand-wavey to me).

Next, we are guaranteed the existence of $\sup E$ by the completeness of $\mathbb{R}$, since $E$ is a subset of $\mathbb{R}$ that is bounded above. Call this $y = \sup E$.

Now, suppose $k \geq 1$.

  1. Assume $y^{2} < k$. I am trying to find an $\epsilon > 0$ such that $(y+\epsilon)^{2} < k$ in order to rule out this possibility. I figured that I would begin by playing around with $(y+\epsilon)^{2}$ in order to put some kind of bounds on it, thereby figuring out what $\epsilon$ should be. I got up to the following point:

$(y+\epsilon)^{2} = (y+\epsilon)(y+\epsilon) = y^{2} + 2\epsilon y + \epsilon^{2}$. Now, since $\epsilon$ is small, I can assume $\epsilon \leq 1$ (can I?), so I get that

$y^{2} + 2\epsilon y + \epsilon^{2} \leq y^{2} + 2\epsilon y + \epsilon = y^{2} + \epsilon(2y + 1)$, but then I got stuck.

I noticed this problem is similar to the proof of Theorem 1.21 in Baby Rudin (for the case where $n = 2$), and I see that he got more bounds, but I'm not entirely sure where they came from. Digging around on math.stackexchange, I found a question where somebody asked where his bounds came from, and the best answer was something along the lines of Newton's method, which I can't use. Anything I use to help me prove this needs to be something easily proveable from the field properties of $\mathbb{R}$ or the order properties of $\mathbb{R}$.

If someone could please help me proceed with this problem, it would be much appreciated!

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  • you know that $x\mapsto x^2 $ is strictly monotonically increasing for $x>0$? – Thomas Sep 07 '15 at 17:44
  • We have proven a result to that effect. How does that help? –  Sep 07 '15 at 17:45
  • it immediately implies uniqueness. If $x\neq y$ then either $x < y$ or $y < x$ so a similar inequality holds for the squares, which means they are not equal. Then it allows you to fix the $k< 1$ case for boundedness. It also simplifies your quest to show that $y^2 = k$, since you immediately arrive at a contradiction by monotonicity if this is wrong... – Thomas Sep 07 '15 at 17:59
  • Okay, say I didn't want to go that route (I was actually instructed to do the proof a certain way), how can I bound $y^{2} + \epsilon(2y+1)$ in a way that will allow me to choose my $\epsilon$ in order to get a contradiction? –  Sep 07 '15 at 18:05
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    Ok, I see my last comment was not too helpful, so I deleted it. If $y^2 < k$, then let $\delta = k- y^2$ You want $y\varepsilon + \epsilon^2 < \delta$, so choose a natural number $n$ such that $n>y$ and choose $\varepsilon <\min { \delta/(n+1), 1}$. Then $y\varepsilon + \epsilon^2 \le (y+1)\varepsilon <(n+1)\varepsilon < \delta$ . – Thomas Sep 07 '15 at 18:21
  • It just warms my cockles when people down vote, don't give a reason, and even if they did, it would probably be a pretty lame one. –  Dec 21 '15 at 04:30

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