# Equivalent characterization of uniform convergence

1. May 7, 2007

### twoflower

Hi all,

I'm learning some calculus theory and I found one point I don't fully understand:

$$\mbox{Let M} \subset \mathbb{R} \mbox{ be non-empty set and let } f, f_{n}, n \in \mathbb{N} \mbox{ be functions defined on M. Then the following is true:}$$

$$f_n \rightrightarrows f \mbox{ on M} \Leftrightarrow \lim_{n \rightarrow \infty} \sup \left\{\left|f_{n}(x) - f(x)\right|; x \in M \right\} = 0$$

Proof:

$$f_{n} \rightrightarrows f \mbox{ on M }$$

$$\Leftrightarrow\ \forall \epsilon > 0\ \exists n_{0} \in \mathbb{N}\ \forall n \geq n_{0} \forall x \in M\ : |f_{n}(x) - f(x)| < \epsilon$$

$$\Leftrightarrow \forall \epsilon > 0\ \exists n_{0} \in \mathbb{N}\ \forall n \geq n_{0} : \sup_{x \in M} \left\{ | f_{n}(x) - f(x) | \leq \epsilon \right\}$$

$$\Leftrightarrow \lim_{n \rightarrow \infty} \left( \sup_{x \in M} |f_{n}(x) - f(x)| \right) = 0$$

I don't get why in the last but one condition in the proof there is [itex]\leq \epsilon[/tex] instead of [itex]< \epsilon[/tex].

Could you please tell me the reason?

Thank you very much

Best regards,
Standa.

Last edited: May 7, 2007
2. May 7, 2007

### matt grime

This case it is the standard result that if S is a set of real number and, for all s in S, s<K, then sup(s)<=K. E.g. take S =(0,1), every s in S is strictly less than 1, but the sup is 1.

3. May 7, 2007

### twoflower

Thank you matt! I see it now.