# intro topology proof linking continuity and convergence

• Sep 2nd 2007, 04:24 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, 05:19 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, 06:12 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 $\displaystyle (x_n)$ be convergent sequence of real numbers. And $\displaystyle x_n \geq M$ for sufficiently large $\displaystyle n$ then $\displaystyle \lim \ x_n \geq M$.

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