# Analysis Proof- different conditions.

A continuous function on $[a,b]$ is also uniformly continuous on $[a,b]$.

The following tries to illustrate what happens when the interval is not closed:

Show: $f(x) = \frac{1}{x}$ is not uniformly continuous in the half open interval ($0, 1$]

Proof:

Take $\epsilon = 1$. We show that there is no $\delta>0$ such that $\forall p$ and $\forall x$ such that $|p − x| < \delta$ $|f(p) − f(x)| < \epsilon$ Take sequences $x_n =\frac{1}{n}$ and $y_n =\frac{1}{n+1}$ . Then $|f(x_n) − f(y_n)| = 1$, but $|x_n − y_n| \rightarrow 0$. So for any $\delta > 0$, there exists $n$ such that $|x_n − y_n| < \delta$ but $|f(x_n) − f(y_n)| \not< 1$. So $f$ is not uniformly continuous.

I looked at this proof, followed the definition/theorem chase, was pleased I understood, till I realised: wait... where have we used the fact this is the half open interval ($0,1$]? If I didnt know any better I would have believed this if it was a claimed disproof of the uniform continuity of $f$ on [$0,1$] Could someone please explain?

• well you can't extend $f$ on $[0,1]$ such that $f$ remains continuous. thus, any such $f$ won't be uniformly continuous. Aug 17, 2015 at 19:51

$f$ is not defined at $0$, so it cannot be continuous at zero.