## Tag Archives: continuou

### Some properties of continuous functions : continued

We now turn to the question whether $f(x_{n})$ approximates $f(x)$ when $x_{n}$ approximates x. This is the same as asking: suppose $(x_{n})_{n=1}^{\infty}$ is a sequence of real numbers converging to x, does $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$?

Theorem.

If $f: \Re \rightarrow \Re$ is continuous at $x \in \Re$ if and only if $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$ whenever $(x_{n})_{n=1}^{\infty}$ converges to x, that is,

$\lim_{n \rightarrow \infty}f(x_{n})=f(\lim_{n \rightarrow \infty} x_{n})$.

Proof.

Suppose f is continuous at x and $(x_{n})_{n=1}^{\infty}$ converges to x. By continuity, for every $\varepsilon > 0$, there exists $\delta >0$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. Since $(x_{n})_{n=1}^{\infty}$ converges to x, for this $\delta >0$, we can find an $n_{0}$ such that $|x-x_{0}|<\delta$ for $n > n_{0}$.

So $|f(x)-f(x_{n})|<\varepsilon$ for $n > n_{0}$ as $|x_{n}-x|<\delta$.

Conversely, suppose $(f(x_{n}))_{n=1}^{\infty}$ converges to $f(x)$ when $(x_{n})_{n=1}^{\infty}$ converges to x. We have to show that f is continuous at x. Suppose f is not continuous at x. That is to say, there is an $\varepsilon >0$ such that however small $\delta$ we may choose, there will be a y satisfying $|x-y|<\delta$ yet $|f(x)-f(y)|\geq \varepsilon$. So for every n, let $x_{n}$ be such a number for which $|x-x_{n}|<(1/n)$ and $|f(x)-f(x_{n})| \geq \varepsilon$. Now, we see that the sequence $(x_{n})_{n=1}^{\infty}$ converges to x. But $(f(x_{n}))_{n=1}^{\infty}$ does not converge to f(x) violating our hypothesis. So f must be continuous at x. QED.

More later,

Nalin Pithwa