March 8th, 2017, 10:28 AM | #11 | |||
Senior Member Joined: Aug 2012 Posts: 1,414 Thanks: 342 | Quote:
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Also note that in Chapter 4, Rudin defines continuity for functions in a metric space, and shows (theorem 4.6) that a continuous function preserves limits. Last edited by Maschke; March 8th, 2017 at 10:40 AM. | |||
March 8th, 2017, 10:32 AM | #12 |
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra |
A) You've missed the point described above. B) No convergent sequence in $\mathbb R$ maps to one that "converges" to zero on $(0,1)$ because such sequences in $\mathbb R$ diverge to $-\infty$. |
March 11th, 2017, 12:07 AM | #13 | |
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,079 Thanks: 87 | Quote:
1/n converges to a limit (which is not in (0,1)). f(1/n) converges to a limit (which is in R because R is complete). -> 1/n converges to a limit in (0,1). Contradiction. | |
March 11th, 2017, 05:29 AM | #14 | |
Math Team Joined: Dec 2013 From: Colombia Posts: 6,854 Thanks: 2228 Math Focus: Mainly analysis and algebra | Quote:
Why would you continue to peddle the same fiction? Last edited by skipjack; March 11th, 2017 at 10:30 AM. | |
March 18th, 2017, 07:49 AM | #15 | ||
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,079 Thanks: 87 | Quote:
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