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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.

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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.

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

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