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Sequences and convergence in the standard topology

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moweee
#1
Apr3-12, 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 {an} converges to a in the standard topology, denoted by τs, then {can} converges to ca in the standard topology on ℝ.


Proof

Let c [itex]\in[/itex] ℝ such that c≠0. Suppose {an} converges to a in the standard topology, denoted by τs.

Let V [itex]\in[/itex] τs with ca in V. Since V in τs, 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] τs such that a [itex]\in[/itex] (p/c, q/c).

Since, by our assumption, {an} converges to a in the standard topology, there exists m [itex]\in[/itex] N such that an [itex]\in[/itex] (p/c,q/c) for all n ≥ m. Hence, p/c < an < q/c. Hence, p < can< q for all n≥m. Since can [itex]\in[/itex] (p,q) and (p,q) [itex]\subseteq[/itex] V, can [itex]\in [/itex]V for all n ≥ m.

Therefore, {can} converges to ca in the standard topology on ℝ.
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micromass
#2
Apr3-12, 06:56 AM
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micromass's Avatar
P: 18,240
Looks pretty good. Two minor points.

Quote Quote by moweee View Post
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 {an} converges to a in the standard topology, denoted by τs, then {can} converges to ca in the standard topology on ℝ.


Proof

Let c [itex]\in[/itex] ℝ such that c≠0. Suppose {an} converges to a in the standard topology, denoted by τs.

Let V [itex]\in[/itex] τs with ca in V. Since V in τs, 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.
Beware that c might be negative.

Note that p/c < (p/c + a)/2 < a < (q/c + a)/2 < q/c
I don't really see the point of this line. Why is this necessary??

Thus, (p/c , q/c) [itex]\in[/itex] τs such that a [itex]\in[/itex] (p/c, q/c).

Since, by our assumption, {an} converges to a in the standard topology, there exists m [itex]\in[/itex] N such that an [itex]\in[/itex] (p/c,q/c) for all n ≥ m. Hence, p/c < an < q/c. Hence, p < can< q for all n≥m. Since can [itex]\in[/itex] (p,q) and (p,q) [itex]\subseteq[/itex] V, can [itex]\in [/itex]V for all n ≥ m.

Therefore, {can} converges to ca in the standard topology on ℝ.
moweee
#3
Apr3-12, 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 interval--specifically, 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?

micromass
#4
Apr3-12, 10:47 AM
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micromass's Avatar
P: 18,240
Sequences and convergence in the standard topology

Quote Quote by moweee View Post
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?
That's ok.

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 interval--specifically, 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).
Ok, I see what you mean. But what is your definition of the standard topology?? Can't you just say that (p/c,q/c) is the interval contained in (p/c,q/c)?? Not that it matters much...

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?
The existence follows directly from the definition of convergence, no? I mean: if you know that an converges, then you automatically get the existence of an m such that blablabla.

If 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.
moweee
#5
Apr3-12, 01:13 PM
P: 3
Quote Quote by micromass View Post
Can't you just say that (p/c,q/c) is the interval contained in (p/c,q/c)?? Not that it matters much...
Oh duh! That makes sense and is a lot less complicated.

Quote Quote by micromass View Post
The existence follows directly from the definition of convergence, no? I mean: if you know that an converges, then you automatically get the existence of an m such that blablabla.

If 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.
Thanks for clearing that up. I appreciate your help =]


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