Thread: intro topology proof linking continuity and convergence

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.

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 be convergent sequence of real numbers. And for sufficiently large then .

Proof: Let . That means . Now in sake of a contradiction say . That means , and so choese . This means, . Thus, . We have shown that . Which is a contradiction because we assumed that for sufficiently large which fails heir. Thus, . Q.E.D.