# intro topology proof linking continuity and convergence

• Sep 2nd 2007, 05:24 PM
ml692787
intro topology proof linking continuity and convergence
I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

Thank you.
• Sep 2nd 2007, 06:19 PM
JakeD
Quote:

Originally Posted by ml692787
I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

Thank you.

For what it's worth, I cannot see a straightforward proof either. At some point you must say, "If a < 0", and show that leads to x_n < 0, contradicting a hypothesis.
• Sep 2nd 2007, 07:12 PM
ThePerfectHacker
Quote:

Originally Posted by ml692787
I'm trying to figure out how to prove this theorem straightforward, i have figured out the proof by contradiction letting a < 0 and epsilon being -a/2, but can't seem to get it straightforward, here is the proof:

if a sequence of numbers (x_n) approaches "a", and the sequence of numbers (x_n) is larger or equal to 0, then "a" is larger than or equal to 0.

Thank you.

Theorem: Let $(x_n)$ be convergent sequence of real numbers. And $x_n \geq M$ for sufficiently large $n$ then $\lim \ x_n \geq M$.

Proof: Let $L = \lim x_n$. That means $|x_n - L| < \epsilon \mbox{ for }n\geq N\in \mathbb{N}$. Now in sake of a contradiction say $L. That means $\frac{M-L}{2}>0$, and so choese $\epsilon = \frac{M-L}{2}$. This means, $|x_n - L | < \frac{M-L}{2} \mbox{ for }n\geq N$. Thus, $x_n -L < \frac{M-L}{2} \implies x_n < \frac{M-L}{2}+L = \frac{M+L}{2} < \frac{M+M}{2} = M$. We have shown that $x_n < M \mbox{ for }n\geq N$. Which is a contradiction because we assumed that $x_n \geq M$ for sufficiently large $n$ which fails heir. Thus, $\lim \ x_n = L \geq M$. Q.E.D.