*Notes to a video lecture on http://www.unizor.com*

__Sequence Limit -__

Bounded Sequence

Bounded Sequence

A very important property of bounded sequences (those

**{**that can be "framed" in upper and lower bounds as

*x*}_{n}

*A ≤ x*_{n}≤ B**) is the following theorem.**

*n**Bolzano - Weierstrass Theorem*

**Any bounded sequence has a convergent subsequence.**

*Proof*

Intuitively, an infinite sequence in a bounded space must have points of accumulation and, therefore, we can pick subsequence that converges to one of these points.

Rigorous proof requires some more precision.

Let's use the method of nested intervals.

Since

*A ≤ x*_{n}≤ B**, we can state that infinite number of elements of our sequence are in the interval**

*n***.**

*I*=[_{1}*A,B*]Split our interval

**in two equal parts at midpoint**

*I*_{1}**. Since original interval**

*M*_{1}**[**has infinite number of members of our sequence, one of its halves or both must also contain an infinite number of them. Assume for definiteness that interval

*A,B*]**is the one with infinite number of members of our sequence. Note that interval**

*I*=[_{2}*A,M*]_{1}**is a subset of interval**

*I*_{2}**:**

*I*_{1}**⊃**

*I*_{1}**.**

*I*_{2}Split our interval

**in two equal parts at midpoint**

*I*_{2}**. Since the "parent" interval**

*M*_{2}**[**has infinite number of members of our sequence, one of its halves or both must also contain an infinite number of them. Assume for definiteness that interval

*A,M*]_{1}**is the one with infinite number of members of our sequence. Note that interval**

*I*=[_{3}*M*]_{2},M_{1}**is a subset of interval**

*I*_{3}**:**

*I*_{2}**⊃**

*I*_{2}**.**

*I*_{3}This process of splitting intervals produces an infinite sequence of nested intervals:

**⊃**

*I*_{1}**⊃**

*I*_{2}**⊃...⊃**

*I*_{3}**⊃**

*I*_{k}**⊃...**

*I*_{k+1}Each subsequent interval is half the size of a previous interval and infinite number of members of our original sequence

**{**is located in each of them.

*x*}_{n}Let's choose any one point

**from each**

*y*_{k}**. We will prove that this subsequence**

*I*_{k}**{**of the original sequence

*y*}_{k}**{**converges to some point inside interval

*x*}_{n}**[**.

*A,B*]Consider the left end of each of these intervals. Since intervals are nested, these left ends produce the monotonically increasing sequence of real numbers bounded from above by point

**and, therefore, have a limit - point**

*B***. This had been proven in "Algebra - Limits - Theoretical Problems" lecture of this course.**

*L*_{1}Similarly, consider the right end of each of these intervals. Since intervals are nested, these right ends produce the monotonically decreasing sequence of real numbers bounded from below by point

**and, therefore, have a limit - point**

*A***.**

*L*_{2}Since the length of our nested intervals converges to zero, it is obvious that points

**and**

*L*_{1}**coincide. Let's call this one point**

*L*_{2}**.**

*L*This point

**is the also the limit of our subsequence**

*L***{**because this subsequence is bounded from below by left ends of our nested intervals, bounded from above by their right ends, and these two bounding sequences of left and right ends converge to the same limit

*y*}_{k}**. This had been proven in "Algebra - Limits - Theoretical Problems" lecture of this course.**

*L*End of proof.

## No comments:

Post a Comment