0
$\begingroup$

Hello i need to know what is the exact meaning of this expression: Continuity and Uniform Continuity for functions of one variable (topological definitions) in analysis course. Does it mean the usual definition:

If $f:D\rightarrow \mathbb{R}$ is a function and $a\in D$ then we say that f is continuous at a if: $(\forall \epsilon>0) (\exists \delta>0) (\forall x\in D) , |x-a|< \delta \rightarrow |f(x)-f(a)|<\epsilon$.

Or it's the same definition but using metric distances instead of absolute values.

$\endgroup$
0
$\begingroup$

You should use distance on the left and absolute value on the right, that is,$$d(x,a)<\delta\implies\bigl|f(x)-f(a)\bigr|<\varepsilon.$$

$\endgroup$
  • $\begingroup$ If D\subset \mathbb{R}, then this distance at left would be the absolute value? $\endgroup$ – Rabih Assaf Apr 6 '18 at 11:06
0
$\begingroup$

The definition of continuity at a point is

If $$f:D→R$$ is a function and $a∈D$ then we say that $f$ is continuous at $a$ if: $$(∀ϵ>0)(∃δ>0)(∀x∈D),d(x,y)<δ \implies |f(x)−f(a)|<\epsilon .$$

A function is continuous on $D$ if it is continuous at every point $x\in D$

On the other hand the uniform continuity is defined as

If $$f:D→R$$ is a function, then we say that $f$ is uniformly continuous on $D$

if: $$(∀ϵ>0)(∃δ>0)(∀x , y ∈D),d(x,y)<δ\implies |f(x)−f(y)|<\epsilon .$$

$\endgroup$
  • $\begingroup$ Hello, are these definitions considered as topological definitions? $\endgroup$ – Rabih Assaf Apr 6 '18 at 10:05
  • $\begingroup$ Yes, in standard topology of real line these are topological definitions. If you use a different metric then the absolute values change to d(x,y). $\endgroup$ – Mohammad Riazi-Kermani Apr 6 '18 at 10:08
  • $\begingroup$ @RabihAssaf they are the standard "metric definitions", we could replace the open balls in the domain by relatively open neighbourhoods, if you prefer. Uniform continuity needs a metric (or really, a uniformity) on the domain. There is no purely topological alternative, of course. $\endgroup$ – Henno Brandsma Apr 7 '18 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.