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

 

 

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