Proof that T is bounded below with ##inf T = 2M##

ChiralSuperfields · Mar 11, 2024

My first solution is

Let
##S = \{x_1, x_2, x_3, ..., x_n\}##
##T = \{2x_1, 2x_2, 2x_3, ... 2x_n\}##
##T = 2S##

Therefore, ##inf T = inf 2S = 2inf S = 2M##

May someone please know whether this counts as a proof?

My second solution is,

##x ≥ M##
##2x ≥ 2M##
##y ≥ 2M## (Letting y = 2M)

Let ##inf T = N##
Therefore by using definition of infimum,
##N ≥ K## where ##K## is a lower bound
Let ##K = \frac{x}{n}## where ##n > 1##
Suppose ##N = 2M##
##N ≥ K##
##N ≥ \frac{x}{n}##
##2M ≥ \frac{x}{n}##

However, I am unsure where to go from here.

Any help greatly appreciated - Chiral.

pasmith · Mar 12, 2024

The first proof is invalid, because it is not given that [itex]S[/itex] is finite (or countable).

Your aim is to show that [itex]2M = 2\inf S[/itex] is the infimum of [itex]T = \{ 2x: x \in S\}[/itex]. Do it directly: is [itex]2M[/itex] a lower bound for [itex]T[/itex]? If [itex]b > 2M[/itex], is [itex]b[/itex] a lower bound for [itex]T[/itex]?

docnet · Mar 14, 2024

Your goal is to show that for all ##t\in T##, ##2M\leq t##.

WWGD · Mar 14, 2024

ChiralSuperfields said:

Homework Statement: I have found two alternative proofs to the solutions of the problem below so I am wondering whether my proofs are valid.
Relevant Equations: ##inf T = 2M##

View attachment 341650
View attachment 341651
My first solution is

Let
##S = \{x_1, x_2, x_3, ..., x_n\}##
##T = \{2x_1, 2x_2, 2x_3, ... 2x_n\}##
##T = 2S##

Therefore, ##inf T = inf 2S = 2inf S = 2M##

May someone please know whether this counts as a proof?

My second solution is,

##x ≥ M##
##2x ≥ 2M##
##y ≥ 2M## (Letting y = 2M)

Let ##inf T = N##
Therefore by using definition of infimum,
##N ≥ K## where ##K## is a lower bound
Let ##K = \frac{x}{n}## where ##n > 1##
Suppose ##N = 2M##
##N ≥ K##
##N ≥ \frac{x}{n}##
##2M ≥ \frac{x}{n}##

However, I am unsure where to go from here.

Any help greatly appreciated - Chiral.

For one, whether ##2x >x## will depend on the sign of ##x##. ##8>4##, but not so for ##-2, -1##.

nuuskur · Saturday, 1:05 PM

The first argument does not count since that's what you have to prove in the first place. Why can you pull the ##2## in front of the infimum?

Either way ##2M## is clearly a lower bound for ##T## and by assumption we can find ## x\in S ## satisfying ##x<M+\frac{\varepsilon}{2}##, hence ##T\ni 2x < 2M+\varepsilon ##.

Proof that T is bounded below with ##inf T = 2M##

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