Math 394 - Topology
Solutions to Homework due February 5
Question 1. Below are listed several sequences in R2with the Euclidean metric. For each of the
sequences, (i) sketch the sequence in R2, (ii) decide if you think that the sequence is bounded, and
(iii) decide if you think the the sequence converges. For (i), by far the best way to plot these is to
plug in several values of n, like n= 1,2,3,4,5,6,7,8 and get an idea for what the sequence looks
like. For (ii), if you think it is bounded, find an a∈R2and M > 0 such that xn∈BM(a) for all n.
For (iii), if you think it converges, to what do you think it converges? [No formal proof is necessary]
(a) xn=1,1
nfor n≥1.
(b) xn=1
n,(−1)nfor n≥1.
(c) xn=cos πn
2,sin πn
2 for n≥0.
(d) xn=ncos πn
2, n sin πn
2 for n≥0.
Solution 1a. This sequence is bounded by using a= (1,0) and M= 2. This sequence converges
to (1,0).
Solution 1b. This sequence is bounded by using a= (0,0) and M= 2. This sequence diverges
(does not converge) since it bounces around too much.
Solution 1c. This sequence is bounded by using a= (0,0) and M= 2. This sequence does not
converge. It seems to cycle amongst the same four points.
Solution 1d. This sequence is not bounded. It is also divergent.
Question 2. As in class, show that the sequence xn=1
n2defined for n≥1 converges to 0 in the
metric space Rwith the standard Euclidean metric.
Solution 2. Given ε > 0, consider N=1
√ε. Notice that since ε > 0, our Nis well-defined. For
n > N, we have
n > 1
√ε.
Since n > 0 and √ε > 0, we cross multiply to obtain
1
n<√ε.
Square both sides of an inequality maintains the inequality (since x2is an increasing function for
positive values). Thus, we obtain 1
n2< ε.
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