Looks pretty good. Two minor points.
Quote by 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 [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.

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] 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.
