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micromass
#2
Apr3-12, 06:56 AM
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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.