Uniformly cauchy sequencesA sequence of functions f n is said to be uniformly cauchy if ∀ ε &gt; 0 ∃ N &gt; 0 : ∀ z , ∀ r , s &gt; N : | f r ( z ) − f s ( z ) | &lt; εHow can I show that if a sequence is uniformly cauchy then f n converge uniformly to some funciton f? We can assume that the metric space is complete.

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Uniformly cauchy sequencesA sequence of functions       f    n   is said to be uniformly cauchy if  ∀  ε  &amp;gt;  0     ∃  N  &amp;gt;  0  :  ∀  z  ,  ∀  r  ,  s  &amp;gt;  N  :      |        f    r    (  z  )  −      f    s    (  z  )      |    &amp;lt;  εHow can I show that if a sequence is uniformly cauchy then       f    n   converge uniformly to some funciton   f? We can assume that the metric space is complete.

Alexander Sweeney · Accepted Answer

Here is how to avoid the &quot;let   m  →  ∞&quot; (see above). Let   f be the pointwise limit (see above). If   ϵ  &amp;gt;  0  , then so is   ϵ      /    2  &amp;gt;  0 hence there exists a natural number       N          ϵ              /            2       such that  ∀  z  ,  ∀  n  ,  m  &amp;gt;      N          ϵ              /            2        :      |          f              n              (    z    )    −          f              m              (    z    )    |    &amp;lt;            ϵ      2      Let   z be arbitrary. If   n  &amp;gt;      N          ϵ              /            2      , then we have      |    f    (    z    )    −          f              n              (    z    )    |    ⩽      |    f    (    z    )    −          f              m              (    z    )    |    +      |          f              m              (    z    )    −          f              n              (    z    )    |  This is true for every   m. In particular       |    f    (    z    )    −          f              m              (    z    )    |    &amp;lt;            ϵ      2       if   m  &amp;gt;  l  (  z  ), where   l  (  z  ) is a natural number dependent on   z, because   f is the pointwise limit. Hence, if   m  &amp;gt;  max      {          N              ϵ                  /                2              ,    l    (    z    )    }  , we get      |    f    (    z    )    −          f              n              (    z    )    |    &amp;lt;  ϵWe have thus shown that for every   ϵ  &amp;gt;  0 there exists an   N, independent of   z, such that      |    f    (    z    )    −          f              n              (    z    )    |    &amp;lt;  ϵ  ,  ∀  n  &amp;gt;  N  ,which is exactly what we needed.

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