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The definition of continuity is:

$f$ is continuous at $a$ if: Given any $\epsilon>0 $,

$\exists \delta > 0$ st. $|x-a|<\delta \implies |f(x)-f(a)|< \epsilon$

$\delta$ obviously depends on the given $\epsilon $ ($\delta=\delta(\epsilon))$ and the range of values $x $ can take depends on $\delta$ but,

am I right in thinking that $\delta$ does not depend on $x$?

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  • $\begingroup$ For fixed $f$ it depends on $\epsilon$ and $a$. $\endgroup$ – André Nicolas Jan 26 '15 at 22:30
  • $\begingroup$ Yes, you are. What we mean by writing |x - a| < delta is that x is a variable in a small ball of radius delta centred at a. So there's no reason to regard delta as dependent on x. $\endgroup$ – user207710 Jan 26 '15 at 22:32
  • $\begingroup$ Your confusion reflects the fact that the continuity at $a$ of $f$ defined on $U$ should be defined as follows: $$\forall\varepsilon\gt0,\ \exists\delta\gt0,\ \forall x\in U,\ |x-a|\lt\delta\implies|f(x)-f(a)|\lt\varepsilon.$$ $\endgroup$ – Did Feb 8 '15 at 17:02
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If $f$ is uniformly continuous, then the choice of $\delta$ is independent of $a$. However, in general, the choice of $\delta$ depends on how the function behaves near $a$.

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