Sequence and Convergence

Convergent Sequence

Def Sequence Let be a set. A sequence in is a function . We usually write and write the sequence as or or .

Subsequence

If is a sequence, a subsequence of is a sequence , where is a strictly increasing sequence in .

Def Convergence A sequence in metric space converges to if In terms of open balls, this can be phrased as for every , exists such that Lemma A sequence in a metric space can have at most one limit.

Def Bounded Sequence Let be a metric space. A sequence in is called bounded if the subset of consisting of all elements from is bounded, i.e. there exist and such that for all .

Thrm In metric spaces convergent sequences must be bounded. Proof Assume . Then there exists such that for all . Let . Then for all and hence for all . The boundedness of thus follows.

Thrm In any metric space we have Proof

Thrm Let be a normed vector space on . If and in and in , then

Thrm Let be an inner product space. If and , then . Proof We have Therefore, by the Cauchy-Schwarz inequality, we have because , and is bounded.

Convergence and Closeness

Thrm Let be a sequence in a metric space . We have , as if and only if for every open set containing there is an such that for all . Proof

Thrm Let be a metric space and . Then Proof We have Thus proved the proposition.

Corollary Let be a metric space and . Then is closed iff the limit of every convergent sequence in is still in . Proof Assume is closed. If is a sequence in with , then we have . Since is closed, and hence . Conversely, assume every convergent sequence in has its limit in . If , by theorem there is a sequence in such that . Thus by the given assumption. This means that and hence . That is is closed.

Thrm Every sequence in has a monotone subsequence. Proof Let be a sequence in . Let . Then Then is a subset of which can be finite or infinite. If is infinite, say , then is a subsequence of that is decreasing. If is finite, there is such that for any . Take . Since , there is such that xn1 <xn2. Since n2 ̸∈S, there is n3 >n2 such that xn2 < xn3 . By repeating this procedure, we can obtain a subsequence (xnj ) of (xn) that is strictly increasing.

Thrm Bolzano-Weierstrass Theorem Every bounded sequence in has a convergent subsequence.