sequences and convergence in the standard topologyby moweee Tags: convergence, sequences, standard, topology 

#1
Apr312, 04:48 AM

P: 3

Hello all.
I have to present a proof to our Intro to Topology class and I just wanted to make sure I did it right (before I look like a fool up there). Proposition Let c be in ℝ such that c≠0. Prove that if {a_{n}} converges to a in the standard topology, denoted by τ_{s}, then {ca_{n}} converges to ca in the standard topology on ℝ. Proof Let c [itex]\in[/itex] ℝ such that c≠0. Suppose {a_{n}} converges to a in the standard topology, denoted by τ_{s}. Let V [itex]\in[/itex] τ_{s} with ca in V. Since V in τ_{s}, there exists an interval (p,q) with ca [itex]\in[/itex] (p,q) and (p,q) [itex]\subseteq[/itex] V. Thus, p < ca < q, which implies p/c < a < q/c. Note that p/c < (p/c + a)/2 < a < (q/c + a)/2 < q/c Thus, (p/c , q/c) [itex]\in[/itex] τ_{s} such that a [itex]\in[/itex] (p/c, q/c). Since, by our assumption, {a_{n}} converges to a in the standard topology, there exists m [itex]\in[/itex] N such that a_{n} [itex]\in[/itex] (p/c,q/c) for all n ≥ m. Hence, p/c < a_{n} < q/c. Hence, p < ca_{n}< q for all n≥m. Since ca_{n} [itex]\in[/itex] (p,q) and (p,q) [itex]\subseteq[/itex] V, ca_{n} [itex]\in [/itex]V for all n ≥ m. Therefore, {ca_{n}} converges to ca in the standard topology on ℝ. 



#2
Apr312, 06:56 AM

Mentor
P: 16,591

Looks pretty good. Two minor points.




#3
Apr312, 07:45 AM

P: 3

Oh ok. So if I consider cases with c>0 and c<0 then most of the work stays in tact. But in the case where c<0, when I divide by c, the inequalities will change, but this will later be undone when I multiply by c at the end. So it should still work. Correct?
I mentioned that p/c < (p/c + a)/2 < a < (q/c + a)/2 < q/c because I needed to justify that (p/c, q/c) is in fact in the standard topology by showing that there exists an intervalspecifically, the interval ( (p/c + a)/2, (q/c + a)/2 ) such that a is in said interval and said interval is contained in (p/c,q/c). One thing is bothering me though...my prof always stressed that when prove that something "exists" when have to choose/set it. I get confused when using the definition of convergence since it implies that there exists an m\in N such that blah blah blah. By showing such an m exists, is that enough? Or do I have to find a specific m\in N where it works? Am I making any sense? 



#4
Apr312, 10:47 AM

Mentor
P: 16,591

sequences and convergence in the standard topologyIf you want to show convergence, then you need to find a specific m. But here you're already given convergence, so existence is no problem here. 



#5
Apr312, 01:13 PM

P: 3




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