- #1
happyg1
- 308
- 0
Hi,
Here's the question:
Show that if [tex]{x_n}[/tex] is a cauchy sequence of points in the metric space M, and if [tex]{x_n}[/tex] has a subsequence which converges to [tex]x \in M[/tex], Prove that [tex] x_n[/tex] itself is convergent to x.
Now, I have proved this as follows..I didn't put in all of the details...
Let [tex]{x_n_k}[/tex] be the subsequence which converges to x.
Choose [tex]n\in\mathbb{N}[/tex] such that [tex]\forall k \geq N [/tex] the distance from [tex]x_n_k[/tex] to x [tex]\leq\frac{\eps}{2}[/tex] and similarly for [tex]x_n,x_m[/tex] then you wind up with [tex]\frac{\eps}{2}+\frac{\eps}{2}=\eps so you're done.
My confusion lies in why can't you do a proof by contradiction?
You let [tex]x_n[/tex] converge to, say y, and the subsequence [tex]s_n_k[/tex] (by hypothesis) converges to x...but every subsequence of a convergent (cauchy) sequence converges to the same limit. Why doesn't this work?
CC
Here's the question:
Show that if [tex]{x_n}[/tex] is a cauchy sequence of points in the metric space M, and if [tex]{x_n}[/tex] has a subsequence which converges to [tex]x \in M[/tex], Prove that [tex] x_n[/tex] itself is convergent to x.
Now, I have proved this as follows..I didn't put in all of the details...
Let [tex]{x_n_k}[/tex] be the subsequence which converges to x.
Choose [tex]n\in\mathbb{N}[/tex] such that [tex]\forall k \geq N [/tex] the distance from [tex]x_n_k[/tex] to x [tex]\leq\frac{\eps}{2}[/tex] and similarly for [tex]x_n,x_m[/tex] then you wind up with [tex]\frac{\eps}{2}+\frac{\eps}{2}=\eps so you're done.
My confusion lies in why can't you do a proof by contradiction?
You let [tex]x_n[/tex] converge to, say y, and the subsequence [tex]s_n_k[/tex] (by hypothesis) converges to x...but every subsequence of a convergent (cauchy) sequence converges to the same limit. Why doesn't this work?
CC