# Thread: intro topology proof linking continuity and convergence

1. ## 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.

2. 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.

3. 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.

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