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xZero's maths question

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kamil9876:
A lower bound for a set is a number such that for all .

So in particular if is a lower bound, so is every number smaller than . Hence at least one lower bound implies infinitely many.

What you are confusing is "lower bound" and "greatest lower bound" a.k.a , this is unique (and exists if the set is bounded from below by basic properties of ). There can be many lower bounds but there is only (at most) one greatest lower bound.

xZero:
Another question on real analysis,

Let f:[0,1] -> [0,1] be a non increasing, non-continuous function, i.e. whenever . Prove that there exists such that f(c)+c=1. Hint: Let and define c=inf A.

I'm really confused with this question, the first step would be proving that inf A exists, since the set A is both upper and lower bounded (0 and 1), by axiom of completeness, inf A and sup A also exists. Next would be proving that inf A is in the set A, since A is in a inclusive interval, inf A is also in A (not sure if that's enough of a proof).

Now I'm stuck at this point, I have no idea what to do from here on, any hints/examples will be highly appreciated. Thanks

Mao:
I'm not quite sure about the proper notation (I've never paid any attention to this kind of rigor, because applied maths major), but I believe the logic will suffice.

To show that and , it is sufficient to show that and intersects in the domain

Since spans the entire range , and by definition, there must be an intersection.

xZero:
Opps, sorry but I completely missed the part that f(x) is non-continuous, don't think the proof works for a non-continuous function. Thanks tho :(

xZero:
turns out there are parts to this question on the other side of the sheet, which I completely missed, again -.- the questions are kinda confusing so if anyone can help me with some of them that would be awesome.

a)Explain why c exists.
Since X is bounded below, by axiom of completeness, inf X exists, since A is a subset of X, then inf A also exists.

b)Let be a sequence of elements of A that converges to c. Prove that, for all n,
By definition of infimum, for all , and from the non-increasing function, , since

*This is where I got confused, if c is inf A then the sequence of x_n should be decreasing, meaning that x_n > c, but in the later parts it assumes that c>x_n.

c)Deduce that
since x is within [0,1], a closed interval, min X exists, so min A also exists. In this case min A = inf A = c, so c is within A, thus satisfying the condition that

d)If c=0, prove that
Let c=0 since f:[0,1] ->[0,1], the value of f(c) must exist between 0 and 1, thus

e)If c>0, by considering , or otherwise, prove that

This is the most confusing part, from b I assumed the sequence to be decreasing, but in this part it says that x_n<c, so c is no longer inf A. Maybe I misunderstood the original question or something but I've been struggling for ages to figure this one out so can someone please give me some sort of hint?

Thanks

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