Show there's a sequence whose limit is its infimum

[zoxee]

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

 

[zoxee]

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?

 

[LCKurtz]

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 $x_n\rightarrow I$.

 

[zoxee]

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]

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?

Probably so, and yes, you could. Just to make it clear, show me below how you would finish the above argument.

 

[gopher_p]

Make sure your proof works for $\mathbb{N}$; i.e. take into consideration that $\inf S$ may not be a limit point of $S\setminus\{\inf S\}$.

 

[zoxee]

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

 

[zoxee]

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

 

[LCKurtz]

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.

 

[LCKurtz]

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.