shut wave
I've bought an real analysis problem book and wanted to start solving the problems. First problem in the book is about the supremum of a set. I attach my work here, but what made me concerned is the length of the answer at the end of the book to that problem. It is like 2x times mine. Am I missing something?
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I've bought an real analysis problem book and wanted to start solving the proble...
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fickle granite
I've bought an real analysis problem book and wanted to start solving the proble...
Wait, why is x > sqrt(2) - epsilon?
shut wave
well from the definition of supremum we want an x to exist s.t x > S - epsilon, where S is the upper bound of set M. So we have x > sqrt2 - epsilon
fickle granite
well from the definition of supremum we want an x to exist s.t x > S - epsilon, ...
...so then you're assuming sqrt(2) is the supremum, rather than proving it.
Or, wait.
shut wave
Well, to call S the supremum of M, S has to be the upper bound for m, and S-epsilon has to be no upper bound for M
am i wrong
fickle granite
Well, to call S the supremum of M, S has to be the upper bound for m, and S-epsi...
I guess I'm just having trouble reading your work because it's not explicit enough.
Like, I don't think you can just assume that sqrt(2) is an upper bound.
shut wave
why not
shut wave
cant we rewrite the set as (0,sqrt2)∩Q
fickle granite
Maybe we can, maybe we can't.
Look, I'm not saying it's not true that sqrt(2) is an upper bound.
I'm saying you haven't proven it.
shut wave
What's given is `x > 0` and `x^2 < 2`.
ye but solving for x in the second inequality we get that x is in (-sqrt2, sqrt2)
fickle granite
ye but solving for x in the second inequality we get that x is in (-sqrt2, sqrt2...
Right. Solving for x. So not given, proven.
x^2 < 2
sqrt(x^2) < sqrt(2)
|x| < sqrt(2)
x > 0 ==> |x| = x
x < sqrt(2)```And then, once we get that sqrt(2) is an upper bound, we then want to prove that it is a least upper bound.
That is, we want to prove that, for all e > 0, there exists x in M such that x > sqrt(2) - e.
Which I think basically involves proving the density of the rationals in the reals.
fickle granite
I've bought an real analysis problem book and wanted to start solving the proble...
What you wrote is that if there is an x > sqrt(2) - e, then sqrt(2) - e < sqrt(2), which... I don't see how that makes any progress at all.
We know sqrt(2) - e < sqrt(2), because e > 0.
shut wave
*What you wrote* is that *if* there is an `x > sqrt(2) - e`, *then* `sqrt(2) - e...
but doesn't sqrt(2) - e < sqrt(2) mean that left hand side is no upper bound for M
fickle granite
but doesn't sqrt(2) - e < sqrt(2) mean that left hand side is no upper bound for...
No.
For sqrt(2) - e to not be an upper bound, you must prove that there exists x in M such that x > sqrt(2) - e.
That is, if sqrt(2) is the supremum, then sqrt(2) - e < sqrt(2) means sqrt(2) - e is not an upper bound of M, but that's just circular reasoning.
shut wave
sqrt(2) - eps < x < sqrt(2), isn't such an x given by the density
fickle granite
sqrt(2) - eps < x < sqrt(2), isn't such an x given by the density
Density?
shut wave
doesnt it guarantee that such x exists
fickle granite
doesnt it guarantee that such x exists
What's density?
shut wave
For all a,b in R, with a<b, there exists an x in Q such that a < x < b
fickle granite
For all a,b in R, with a<b, there exists an x in Q such that a < x < b
That sounds interesting and really useful. You should prove that.
shut wave
So in order to use the density property of Q, you have to prove it
fickle granite
So in order to use the density property of Q, you have to prove it
I mean, I imagine that's what your book did.
And also you didn't invoke the density.
You just said "sqrt(2) - e < sqrt(2), so sqrt(2) is the least upper bound", which is... not a proof.
shut wave
I mean, I imagine that's what your book did.
completely not
fickle granite
completely not
Show me what your book did.
fickle granite
Okay, it looks like you suppose the supremum is s, you take it as trivial that s > 1, and then you say (s - 1/n)^2 <= 2 <= (s + 1/n)^2?
For natural n?
shut wave
Well, it says that "we can assume s>1"
shut wave
Okay, it looks like you suppose the supremum is s, you take it as trivial that `...
"we will prove that for any natural number n the following inequality holds"
shut wave
english version if you want
fickle granite
english version if you want
See, the part with w, they had to produce a rational number between s + 1/n and s.
shut wave
can we go for contradiction? sfc that sqrt(2) is not the least upper bound. let some k belong to rationals and be the supremum then we have that k < sqrt(2). And by density of Q, there exists an x s.t k < x < sqrt(2). Then, we have that k^2 < x^2 < 2 . We have the x^2 < 2 and also that there is a member of the set that is > than the sup and this is the contradiction
fickle granite
can we go for contradiction? sfc that sqrt(2) is not the least upper bound. le...
Okay, but what if the least upper bound is irrational and less than sqrt(2)?
Okay, look, yes, density gets you your result, but I don't think you get to just assume density.
The book didn't.
If you assume density, the proof is basically two lines. x^2 < 2 ==> x < sqrt(2), therefore sqrt(2) is an upper bound By density, for all x < sqrt(2) there exists y in Q such that x < y < sqrt(2), which means y is in M, which means no x < sqrt(2) can be an upper bound
shut wave
I mean, isn't assuming density one way of doing it and the book way another one?
fickle granite
I really don't think you get to assume density.
shut wave
what's the issue with assuming it?
marble depotBOT
@shut wave
fickle granite
what's the issue with assuming it?
Because, again, the book doesn't assume it.
marble depotBOT
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