Equivalent characterization of uniform convergence

[twoflower]

Hi all,

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

<br /> \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:}<br />

<br /> 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<br />

Proof:

<br /> f_{n} \rightrightarrows f \mbox{ on M }<br />

<br /> \Leftrightarrow\ \forall \epsilon &gt; 0\ \exists n_{0} \in \mathbb{N}\ \forall n \geq n_{0} \forall x \in M\ : |f_{n}(x) - f(x)| &lt; \epsilon<br />

<br /> \Leftrightarrow \forall \epsilon &gt; 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\}<br />

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

I don't get why in the last but one condition in the proof there is \leq \epsilon[/tex] instead of &amp;lt; \epsilon[/tex].&lt;br /&gt; &lt;br /&gt; Could you please tell me the reason?&lt;br /&gt; &lt;br /&gt; Thank you very much&lt;br /&gt; &lt;br /&gt; &lt;br /&gt; Standa.

 

[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.

 

[twoflower]

Thank you matt! I see it now.