Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove if F is continuous on S then for all convergent sequences with $\lim_{k\to \infty}x_k = a$ in s, $\lim_{k \to \infty}f(x_k) = f(a)$

My reasoning: Let $\delta$ = 1/k, then since the norm of $X_k$ and a is less than 1/k and 1/k approaches 0 for k large enough, the sequence converges and is continuous by definition.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

First, sequences are not continuous, I think you meant convergent. Second, why do you say that the norm $|x_k - a|$ is less than $\frac{1}{k}$? This is not justified.

For a limit proof always start by letting $\epsilon > 0$. To conclude that $\lim_{k \to \infty} f(x_k) = f(a)$ we need to show that we can pick a large integer $N$ (that possibly depends on $\epsilon$) such that for all $k > N$ we have $|f(x_k) - f(a)| < \epsilon$.

As $F$ is continuous it is, in particular, continuous at $a$ so by definition there exists a $\delta$ such that for all $x \in S$ we have $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon$. Thus we need to show that we can choose $N$ large such that $k > N$ implies $|x_k - a| < \delta$. But because $\lim_{k \to \infty} x_k = a$ we can, by the definition of a limit, do this.

share|improve this answer
    
Sorry about my sloppyness, i meant to express that way in full –  bobdylan Jan 22 '13 at 19:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.