Does every sequence have a monotone subsequence?

Does every sequence have a monotone subsequence?

It turns out that every sequence of real numbers has subsequence that is monotone.

How do you prove a sequence has a convergent subsequence?

Proof: Every sequence in a closed and bounded subset is bounded, so it has a convergent subsequence, which converges to a point in the set, because the set is closed. Conversely, every bounded sequence is in a closed and bounded set, so it has a convergent subsequence.

What is the statement of monotone subsequence theorem?

Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum.

What sequence has a convergent subsequence?

The theorem states that each bounded sequence in Rn has a convergent subsequence. An equivalent formulation is that a subset of Rn is sequentially compact if and only if it is closed and bounded. The theorem is sometimes called the sequential compactness theorem.

Does every monotone sequence has a convergent subsequence?

Proof. We know that any sequence in R has a monotonic subsequence, and any subsequence of a bounded sequence is clearly bounded, so (sn) has a bounded monotonic subsequence. But every bounded monotonic sequence converges. So (sn) has a convergent subsequence, as required.

Does every convergent sequence have a convergent subsequence?

Every subsequence of a convergent sequence converges to the same limit as the original sequence. if lim sup is finite, then it is the limit of a monotone subsequence. Bolzano-Weierstrass Theorem. Every bounded sequence of real numbers has a convergent subsequence.

Does every monotone sequence have a convergent subsequence?

What is monotone subsequence example?

Monotonicity: The sequence sn is said to be increasing if sn  sn+1 for all n 1, i.e., s1  s2  s3  …. A sequence is said to be monotone if it is either increasing or decreasing. Example. The sequence n2 : 1, 4, 9, 16, 25, 36, 49, is increasing.

Can a divergent monotone sequence have a convergent subsequence?

Since we know every Cauchy sequence is convergent, and every subsequence of a convergent sequence is convergent, this is impos- sible. c) A divergent monotone sequence, with a Cauchy subsequence. Since ak is increasing without bound, the subsequence must be also. Thus, this is impossible.

How to prove that a monotone sequence converges in $BBB R$?

Let $\\langle x_n:n\\in\\Bbb Nangle$ be a bounded, monotone sequence in $\\Bbb R$; without loss of generality assume that the sequence is non-decreasing. To prove that it converges, at some point you’re going to have to use the fact that $\\Bbb R$ is complete: the theorem is not true in $\\Bbb Q$.

How do you find the indices of a monotone subsequence?

An infinite monochromatic subgraph gives us the indices of a monotone subsequence: If red, the subsequence is increasing while, if blue, it is strictly decreasing. Start by noting that there is an infinite A 0 with all edges { 0, i }, i ∈ A 0, of the same color.

How do you know if a set is monochromatic?

Now, the sequence c 0, c 1, c 2, … is a sequence that only takes two values, so it has a constant subsequence. The corresponding i n form the monochromatic set we were looking for. If a n is unbounded, we are done.

How do you prove that a sequence converges?

If you want to prove the statement, if a sequence is monotone and bounded then it converges]