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[quick quest]

Is my proof ok?

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[quick quest]

Proof: Let z be an element of U. Since x is a lower bound of U, xRz. Since z was arbitrary, x is the least upper bound of B.

[glad adder]

which statement are you proving?

@quick quest

[quick quest]

C

[glad adder]

your argument says that x is below all upper bounds of B

but you haven't explained why x is an upper bound of B

@quick quest

in other words, you have used the part where x is a lower bound of U

but you haven't used the assumption that x = inf U

the argument as you have it right now passes for any element of B that is not in U, there's no way to conclude that x = sup B based on that alone

[glad adder]

no

[quick quest]

What does inf and sup mean?

[patent parcel]

infimum and supremum

[glad adder]

infimum = greatest lower bound
supremum = least upper bound

[quick quest]

Could you explain this sentence?

[glad adder]

you said that xRz for any upper bound z of B

but it's not enough

for example B = (0,1), hence U = [1, inf)

take x=0.9, then clearly x is smaller than any upper bound of B

but x is not the supremum

you must show that bRx for all b in B

then your proof is complete

use the fact that x is the infimum of U

@quick quest

[quick quest]

So I need to show x is in U?

[quick quest]

Proof: Since every element of B is a lower bound for U(proved in (b)) and x is the greatest lower bound of U, x is a upper bound of B, which means x is in U. Since x is in U and is a lower bound of U, x is a smallest element in U and therefore x is the least upper bound of B.

@glad adder

[glad adder]

now it's correct

[quick quest]

Thx

[steep ivy]

@quick quest

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[quick quest]

+close

[steep ivy]

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[steep ivy]

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