13 мар. 2016 г. · It's a metric that gives the standard topology. Also, every sequence is bounded with respect to this metric but there are plenty of sequences on ...

30 апр. 2016 г. · Let (xn)∞n=1 be a bounded sequence in E. We claim that X:={xn}∞n=1 is totally bounded. Indeed, suppose this were not the case.

8 апр. 2020 г. · I would do a proof by contradiction: Assume that there's a bounded sequence with no convergent subsequence. Then no point appears infinitely ...

2 авг. 2013 г. · Justify: In a metric space every bounded sequence has a convergent subsequence. My Attempt: False: Consider the metric space (X,d) where X=R ...

16 июл. 2017 г. · I am trying to prove that a bounded sequence in a metric space need not have a convergent subsequence. My counterexample is: Consider the metric ...

4 сент. 2014 г. · It means that each open neighbourhood contains infinitely many pn. So if no subsequence converges to p, there exists an open neighbourhood Vp ...

24 авг. 2019 г. · Let X be a compact metric space. Show that every sequence has a convergent subsequence. Proof (revised version):. Let (xn)n∈N⊆X be a sequence.

6 окт. 2016 г. · It is true that a bounded sequence has a monotonic subsequence, but (i) it need not be increasing, (ii) it need not be strictly monotonic, ...

16 мая 2015 г. · Yes, it's true in every metric space. In fact, a stronger statement is true: every Cauchy sequence is bounded. Let ⟨xn:n∈N⟩ be a Cauchy ...

21 дек. 2019 г. · In a compact metric space every sequence has convergent subsequence but thus is not always true in compact topological spaces. Every sequence in ...