Show there's a sequence whose limit is its infimum

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zoxee
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question:

Suppose ## S \subset \mathbb{R} ## is a nonempty subset of the real numbers that is bounded below.

Show that there exists a sequence ## <x_n> ## such that ## x_n \in S ## for all n and ## \lim (x_n) = inf(S) ##

attempt:

consider an element ## x \in S ## suppose ## x \geq inf(S) + 1/n ## then this would mean that inf(S) is not the greatest lower bound and inf(s) + 1/n is so ## x < inf S + 1/n ## ## \forall n \in \mathbb{N} ## take ## n_1 < n_2 < n_3 < n_4 ... ## then ## x_1 < inf(S) + 1/n_{1} ## and ## x_2 < inf(S) + 1/n_{2} ## and so on so we get a sequence ## inf(S) \leq x_n ... < x_3 < x_2 < x_1 ## hence as n increases 1/n approaches zero (proved already) and x_n approaches inf(S).

Is this all correct as in the solutions all they have stated is:

## inf(S) \leq x_n < inf(S) + 1/n ## and inf(s) + 1/n -> inf(S) so such a sequence exists, I don't understand that
 
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alternatively couldn't I say as ## inf(s) \leq x_n < inf(S) + 1/n ## ##lim inf(S) = inf(S)## and##lim (inf(S) + 1/n) = inf(S) ## therefore by the sandwich theorem x_n approaches inf(S) also?
 
zoxee said:
question:

Suppose ## S \subset \mathbb{R} ## is a nonempty subset of the real numbers that is bounded below.

Show that there exists a sequence ## <x_n> ## such that ## x_n \in S ## for all n and ## \lim (x_n) = inf(S) ##

attempt:

consider an element ## x \in S ## suppose ## x \geq inf(S) + 1/n ## then this would mean that inf(S) is not the greatest lower bound and inf(s) + 1/n is so ## x < inf S + 1/n ## ## \forall n \in \mathbb{N} ## take ## n_1 < n_2 < n_3 < n_4 ... ## then ## x_1 < inf(S) + 1/n_{1} ## and ## x_2 < inf(S) + 1/n_{2} ## and so on so we get a sequence ## inf(S) \leq x_n ... < x_3 < x_2 < x_1 ## hence as n increases 1/n approaches zero (proved already) and x_n approaches inf(S).

Is this all correct as in the solutions all they have stated is:

## inf(S) \leq x_n < inf(S) + 1/n ## and inf(s) + 1/n -> inf(S) so such a sequence exists, I don't understand that

You sort of have the idea but that is a very hard to read and over-complicated explanation. To clean it up, you might start by calling ##I = \inf(S)## so you don't have to keep writing it. Then by the definition of ##\inf(S)## you can say that for each natural number ##n## there is a point ##x_n\in S## in the interval ##(I,I+\frac 1 n)##. Then argue that ##x_n\rightarrow I##.
 
LCKurtz said:
You sort of have the idea but that is a very hard to read and over-complicated explanation. To clean it up, you might start by calling ##I = \inf(S)## so you don't have to keep writing it. Then by the definition of ##\inf(S)## you can say that for each natural number ##n## there is a point ##x_n\in S## in the interval ##(I,I+\frac 1 n)##. Then argue that ##x_n\rightarrow I##.

ok thank you. Is my argument that x_n -> inf(S) valid though? Or could I use the sandwich theorem?
 
LCKurtz said:
You sort of have the idea but that is a very hard to read and over-complicated explanation. To clean it up, you might start by calling ##I = \inf(S)## so you don't have to keep writing it. Then by the definition of ##\inf(S)## you can say that for each natural number ##n## there is a point ##x_n\in S## in the interval ##(I,I+\frac 1 n)##. Then argue that ##x_n\rightarrow I##.

zoxee said:
ok thank you. Is my argument that x_n -> inf(S) valid though? Or could I use the sandwich theorem?

Probably so, and yes, you could. Just to make it clear, show me below how you would finish the above argument.
 
Make sure your proof works for ##\mathbb{N}##; i.e. take into consideration that ##\inf S## may not be a limit point of ##S\setminus\{\inf S\}##.
 
LCKurtz said:
Probably so, and yes, you could. Just to make it clear, show me below how you would finish the above argument.

the limit of I is I and the limit of (I + 1/n) (as 1/n -> 0 (which I've proved previously)) and seeing as x_n is 'sandwiched' by I and I + 1/n the limit of x_n is I, hence a sequence <x_n> which has the infimum as it's limit exists
 
LCKurtz said:
Probably so, and yes, you could. Just to make it clear, show me below how you would finish the above argument.

My other argument would be that if we take ## n_1 < n_2 <n_3 ## and## x_1 < I + 1/n_1## ## x_2 < I + 1/n_2 ## etc then we end up with the sequence ## I \leq x_n <...<x_3<x_2<x_1 < I + 1/n_1 ## and as n approaches infinity then x_n approaches I
 
zoxee said:
the limit of I is I and the limit of (I + 1/n) (as 1/n -> 0 (which I've proved previously)) and seeing as x_n is 'sandwiched' by I and I + 1/n the limit of x_n is I, hence a sequence <x_n> which has the infimum as it's limit exists

Yes. You should probably use the half closed interval ##[I,I+\frac 1 n)## to select your points to address gopher_p's observation.
 
zoxee said:
My other argument would be that if we take ## n_1 < n_2 <n_3 ## and## x_1 < I + 1/n_1## ## x_2 < I + 1/n_2 ## etc then we end up with the sequence ## I \leq x_n <...<x_3<x_2<x_1 < I + 1/n_1 ## and as n approaches infinity then x_n approaches I

While you certainly can make a decreasing sequence ##\{x_n\}##, just picking ##x_k < I+\frac 1 {n_k}## doesn't necessarily do it. But it doesn't need to be decreasing anyway.