Contradictory Proof of Supremum of E: Is it Circular?

[yucheng]

N.B. I have inserted the proof here as reference. See the bolded text.

My question is, isn't the reasoning "$x^{2}+5 \varepsilon<2,$ thus $(x+\varepsilon)^{2}<2 .$ " circular? If we can already find an $0<\varepsilon<1$ such that $x^{2}+5 \varepsilon<2,$ Can't one also claim that " we can find an $0<\varepsilon<1$ such that $(x+\varepsilon)^{2}<2 .$"? If this is the case, why do we need to show the part that $x^{2}+5 \varepsilon<2,$ in the proof below?

Proof. Let $E$ be the set $\left\{y \in \mathbf{R}: y \geq 0\right.$ and $\left.y^{2}<2\right\} ;$ thus $E$ is the set of all non-negative real numbers whose square is less than $2 .$ Observe that $E$ has an upper bound of 2 (because if $y>2,$ then $y^{2}>4>2$ and hence $y \notin E$ ). Also, $E$ is non-empty (for instance, 1 is an element of $E$ ). Thus by the least upper bound property, we have a real number $x:=\sup (E)$ which is the least upper bound of $E$. Then $x$ is greater than or equal to 1 (since $1 \in E$ ) and less than or equal to 2 (since 2 is an upper bound for $E$ ). So $x$ is positive. Now we show that $x^{2}=2$.

We argue this by contradiction. We show that both $x^{2}<2$ and $x^{2}>2$ lead to contradictions. First suppose that $x^{2}<2 .$ Let $0<\varepsilon<1$ be a small number; then we have
$$
(x+\varepsilon)^{2}=x^{2}+2 \varepsilon x+\varepsilon^{2} \leq x^{2}+4 \varepsilon+\varepsilon=x^{2}+5 \varepsilon
$$
since $x \leq 2$ and $\varepsilon^{2} \leq \varepsilon .$ since $x^{2}<2,$ we see that we can choose an $0<\varepsilon<1$ such that $x^{2}+5 \varepsilon<2,$ thus $(x+\varepsilon)^{2}<2 .$ By construction of $E,$ this means that $x+\varepsilon \in E ;$ but this contradicts the fact that $x$ is an upper bound of $E$.

Thanks in advance.

 

[member 587159]

Since $x^2 < 2$ we can choose $0 < \epsilon < 1$ with $x^2 + 5\epsilon < 2$ follows as follows:

Assume to the contrary that $x^2 + 5 \epsilon \geq 2$ for all $\epsilon\in (0,1)$. Rewriting this, we obtain $\epsilon \geq \frac{2-x^2}{5}$ for all $\epsilon \in (0,1)$. Do you see why this is impossible?

 

[yucheng]

Since $x^2 < 2$ we can choose $0 < \epsilon < 1$ with $x^2 + 5\epsilon < 2$ follows as follows:

Assume to the contrary that $x^2 + 5 \epsilon \geq 2$ for all $\epsilon\in (0,1)$. Rewriting this, we obtain $\epsilon \geq \frac{2-x^2}{5}$ for all $\epsilon \in (0,1)$. Do you see why this is impossible?

I mean can I just skip that and go straight to $(x+\epsilon)^2 < 2$? Oh wait, or is it necessary to conclude that $(x+\epsilon)^2 < 2$? Thanks!

 

[member 587159]

I mean can I just skip that and go straight to $(x+\epsilon)^2 < 2$? Oh wait, or is it necessary to conclude that $(x+\epsilon)^2 < 2$? Thanks!

But how would you deduce that? The proof goes through all trouble to derive this. How would you avoid doing this work?

 

[yucheng]

But how would you deduce that? The proof goes through all trouble to derive this. How would you avoid doing this work?

Accidentally short-circuited.

 

[PeroK]

For this proof, I might prefer, assuming $x^2 < 2$, to let $\delta = 2 - x^2$ and work with that. That seems more logical to me.

 

[member 587159]

Accidentally short-circuited.

I guess it tends to happen a lot for beginning analysis students :).

 

[yucheng]

For this proof, I might prefer, assuming $x^2 < 2$, to let $\delta = 2 - x^2$ and work with that. That seems more logical to me.

Do you mean $\delta = 2 - x^2$. After making the approximation for $2x\epsilon + \epsilon^2 < 5\epsilon$, we can find $5\epsilon < \delta$ or $\epsilon < \delta/5$ ? I think it does make things clearer.

 

[PeroK]

Do you mean $\delta = 2 - x^2$. After making the approximation for $2x\epsilon + \epsilon^2 < 5\epsilon$, we can find $5\epsilon < \delta$ or $\epsilon < \delta/5$ ? I think it does make things clearer.

The basis of the proof, as you probably understand, is that:

If $x^2 < 2$, then there must exist a small positive number $\epsilon$ with $(x + \epsilon)^2 < 2$.

One approach is to express $\epsilon$ in terms of $\delta$, which perhaps give you more to work with. Which is what I see you've done.