Sequences and convergence in the standard topology

[moweee]

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 $\in$ ℝ such that c≠0. Suppose {a_n} converges to a in the standard topology, denoted by τ_s.

Let V $\in$ τ_s with ca in V. Since V in τ_s, there exists an interval (p,q) with ca $\in$ (p,q) and (p,q) $\subseteq$ 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) $\in$ τ_s such that a $\in$ (p/c, q/c).

Since, by our assumption, {a_n} converges to a in the standard topology, there exists m $\in$ N such that a_n $\in$ (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 $\in$ (p,q) and (p,q) $\subseteq$ V, ca_n $\in$V for all n ≥ m.

Therefore, {ca_n} converges to ca in the standard topology on ℝ.

 

[micromass]

Looks pretty good. Two minor points.

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 $\in$ ℝ such that c≠0. Suppose {a_n} converges to a in the standard topology, denoted by τ_s.

Let V $\in$ τ_s with ca in V. Since V in τ_s, there exists an interval (p,q) with ca $\in$ (p,q) and (p,q) $\subseteq$ 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) $\in$ τ_s such that a $\in$ (p/c, q/c).

Since, by our assumption, {a_n} converges to a in the standard topology, there exists m $\in$ N such that a_n $\in$ (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 $\in$ (p,q) and (p,q) $\subseteq$ V, ca_n $\in$V for all n ≥ m.

Therefore, {ca_n} converges to ca in the standard topology on ℝ.

 

[moweee]

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]

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 a_n 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]

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.

The existence follows directly from the definition of convergence, no? I mean: if you know that a_n 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 =]