1. ## continuity

Let $\displaystyle f:A \subset \mathbb{R} \to \mathbb{R}$ and $\displaystyle r_n > 0$
a sequence such that $\displaystyle r_n \to 0$
Prove that f is continuos in $\displaystyle \overline x$
if and only if

$\displaystyle s_n = \sup \left\{ {\left| {f\left( x \right) - f\left( {\overline x } \right)} \right|:\,\left| {x - \overline x } \right| \leqslant r_n } \right\}$ converges to zero

thanks!

2. Where are you stuck ?

3. Use this fact, which is "easier" to prove (though more or less the same).

$\displaystyle f:X\to{Y}$ is continuous at $\displaystyle x$ if and only if for every sequence $\displaystyle \{x_n\}\in X$ such that $\displaystyle x_n\to x$, $\displaystyle f(x_n)\to f(x)$.

4. Originally Posted by putnam120
Use this fact, which is "easier" to prove (though more or less the same).

$\displaystyle f:X\to{Y}$ is continuous at $\displaystyle x$ if and only if for every sequence $\displaystyle \{x_n\}\in X$ such that $\displaystyle x_n\to x$, $\displaystyle f(x_n)\to f(x)$.
Careful there, this is only an if and only if theorem if X is metrizable.

In general it is only true that a function being continuous implies for every convergent sequence $\displaystyle x_n \rightarrow x$ implies $\displaystyle f(x_n)\rightarrow f(x)$.

Obviously $\displaystyle \mathbb{R}$ is metrizable, so it is okay to use here, but just want to make sure no passerby thinks this theorem is true in general as you have it stated

5. Oh thank. Guess I have just gotten used to always working with metric spaces that I no longer mention it.