Is Q Complete in the Context of the Completeness Axiom?

[Wingeer]

Proof that Q is "not complete"

Hello.
I have recently started taking a course in the foundations of analysis. We started off with the completeness "axiom": Every upper bounded non-empty subset of R has a supremum (least upper bound). The lecturer then went on demonstrating that Q does not satisfy this. He also gave a proof for it, and that is where he lost me a bit.

Let $$E= \{x \in Q: x^2 < 2 \}$$. E is upper bounded, for example by 2. However it has no supremum. Assume that $$s \in Q$$ is a supremum for E. Clearly $$s>0$$. Also $$s \neq 2$$ since 2 has no square root in Q.
If $$s^2 < 2$$, notice that $$(s+t)^2 = s^2 + 2st + t^2 = s^2 + (2s+t)t$$, where $$t \in Q$$.
If t>0 is chosen to be small, $$s^2 +(2s+t)t<2$$. This contradicts our hypothesis that s is a supremum for E.

I am well aware that the proof is not complete (s^2 >2), but I would like to do and comprehend that by myself.
My only problem with this is where he chooses t to be small so that the expression is less than 2. How can one guarantee that it will be? Would anybody care to elaborate?

 

[mXSCNT]

That doesn't seem complete to me. To really prove it I believe you need to find a rational t for each s.

 

[alexfloo]

Here is how that selection is justified:

Since s²<2, we have 2-s²=q is a positive number. There is no smallest positive number (because for positive t, t/2<t). Therefore, we must select t such that (2s+t)t<q. Can you figure it out from there?

 

[Wingeer]

From there it is just a matter of simple algebra. However I am not sure whether I got why there cannot be a smallest positive number and the relation between q and t?
Thanks in advance!

 

[alexfloo]

Suppose t is the smallest positive number. Then t/2 is also positive, right? (positive times positive is positive). But since 1/2<1, it follows that t/2<t. This means that t is not the smallest positive number, which is a contradiction. Did that make sense? The idea is just that we can cut any number in half and get a smaller one.

As for "q," I just picked it as part of my strategy, but there are other ways to do it. In that step of the proof, we're trying to show that there's another number between s² and 2. Since they're not equal, the difference is a positive number. (I called that number "q"). We can then pick a number smaller than q by the above, and then we can just add that number to s². Since it's less than the "space between" s² and 2, the new number stays less than 2.

Did that help?

 

[Wingeer]

Thanks! The second section of your post really opened my eyes. It helped a lot!

 

[mXSCNT]

Sure, it's intuitively clear why you can choose a value of t that's "small enough," but to really prove it, for each value of s and q you need to actually find the (rational) value of t alexfloo mentioned.

Since s2<2, we have 2-s2=q is a positive number. There is no smallest positive number (because for positive t, t/2<t). Therefore, we must select t such that (2s+t)t<q. Can you figure it out from there?

 

[alexfloo]

I disagree mXSCNT. The fact that the limit is 0 as t goes to 0 is a consequence of the fact that sufficiently small t exists. Since they're equivalent, the converse is true also, but the idea of a limit of a function (or even the idea of a function at all) is a more complex concept than is necessary for this proof.

You certainly could prove that first, but it's certainly not necessary. It's a simple matter of noting that (2s+t)t>2st.

Since 2s is a positive constant with respect to t, taking t<1/(2s) (which again must exist, since 1/(2s) is not the smallest positive real number) is sufficient.

EDIT: Scratch this post. I recall seeing you say something somewhere that I now don't see anywhere at all, haha. Apologies.

 

[mXSCNT]

I disagree mXSCNT. The fact that the limit is 0 as t goes to 0 is a consequence of the fact that sufficiently small t exists. Since they're equivalent, the converse is true also, but the idea of a limit of a function (or even the idea of a function at all) is a more complex concept than is necessary for this proof.

You certainly could prove that first, but it's certainly not necessary. It's a simple matter of noting that (2s+t)t>2st.

Since 2s is a positive constant with respect to t, taking t<1/(2s) (which again must exist, since 1/(2s) is not the smallest positive real number) is sufficient.

EDIT: Scratch this post. I recall seeing you say something somewhere that I now don't see anywhere at all, haha. Apologies.

I did edit my post. But there's a mistake in your logic. (2s+t)t > 2st doesn't help you because you're trying to find a small value of (2s+t)t, not 2st. Showing that you can make 2st small through choice of t is of no use. Better to start with t < s for a range of s near sqrt(2) that we're interested in, and therefore (2s+t)t < (2s+s)t. Then you can set t to something like (2-s^2)/(2s+s), so that (2s+t)t < (2s+s)t = 2-s^2 (and then go back and justify t < s for some range of s).

 

[Wingeer]

I am not sure if I am allowed to hijack my own thread to ask another (possibly unrelated) question, but here goes:
My lecturer has his own way of doing things in terms of notation. The other day, he was introducing subsequences. First off:
$$I \subseteq \mathbb{N}$$
A subsequence of $$< x_n >_{n \in I}$$ is a sequence $$< x_n >_{n \in J}$$ where $$J \subseteq I$$

Now, a possibly improper subsequence of $$<x_i>_{i \in I}$$ is $$<x_i>_{i \in J}$$ where $$J \setminus I$$ is finite.
Now here is my problem. As J is a subset of I the intersection of the sets will just be J. Does not that mean that the last equation is just the empty set? I am suspecting that it really should be the other way around, but I feel like I am walking on pudding in this course. And also the same equation is used several times later on.

 

[Wingeer]

Actually. For your convenience I found a note written by the lecturer regarding this topic. This is probably much better than my second hand notes.
http://www.math.ntnu.no/~hanche/notes/diagonal/diagonal.pdf" is the link.

 

[alexfloo]

Here he's using a convention that I've never seen before but that doesn't actually change the underlying math at all for most purposes. J is not a subset of I, it's "almost" a subset of I. Only finitely many of members "screw up" by being from elsewhere. This doesn't matter because:

***changing finitely many (not matter how large the finite number) terms of a sequence does not effect its limit, limit superior, limit inferior, or really any of its interesting properties at all (except maybe its sum, it it is summable).

So, when we're just studying convergence we can allow finitely much of J to not be in I, and our definition it actually for our purposes equivalent to if we restrict J to be a subset of I. The case there the set-difference is empty is the case of the usual definition that I'm more familiar with.

 

[Wingeer]

I get it now. Thanks. I have not been exposed to much topology-ish math, yet. I guess it is kind of a transition.

 

[evagelos]

Hello.
I have recently started taking a course in the foundations of analysis. We started off with the completeness "axiom": Every upper bounded non-empty subset of R has a supremum (least upper bound). The lecturer then went on demonstrating that Q does not satisfy this. He also gave a proof for it, and that is where he lost me a bit.

Let $$E= \{x \in Q: x^2 < 2 \}$$. E is upper bounded, for example by 2. However it has no supremum. Assume that $$s \in Q$$ is a supremum for E. Clearly $$s>0$$. Also $$s \neq 2$$ since 2 has no square root in Q.
If $$s^2 < 2$$, notice that $$(s+t)^2 = s^2 + 2st + t^2 = s^2 + (2s+t)t$$, where $$t \in Q$$.
If t>0 is chosen to be small, $$s^2 +(2s+t)t<2$$. This contradicts our hypothesis that s is a supremum for E.

I am well aware that the proof is not complete (s^2 >2), but I would like to do and comprehend that by myself.
My only problem with this is where he chooses t to be small so that the expression is less than 2. How can one guarantee that it will be? Would anybody care to elaborate?

The following argument is correct w.r.t choosing t.

$s^2<2\rightarrow 2-s^2>0$ ,but also 2s+1>0 ,hence $\frac{2-s^2}{2s+1}>0$.

Now choose t= min{1 ,$\frac{2-s^2}{2s}$}.

Hence $t<\frac{2-s^2}{2s+1}\rightarrow s^2+2st+t<2$ and $(s+t)^2<s^2+2st+t<2$ ,since t<1 (that explains the expression:"choose t to be very small)

Thus (s+t) belongs to E and (s+t)$\leq s\rightarrow t<0$ ,a contradiction