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Let A ⊂ R. Function f : A → R is continuous in point a ∈ A, if there exists ε > 0, so that with every δ > 0

|f(x) − f(a)| < ε, when x ∈ A and |x − a| < δ.

Is function f : R → R, f(x) = x for all x ∈ R, continuous in point 0 when using this false definition?

jte
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    What do you think? – José Carlos Santos Nov 02 '18 at 18:04
  • I think it's not, but I have only gotten in to a solution that if we choose δ = 2ε, we can get |x| < δ = 2ε. I think it doesn't still proof anything as it doesn't say |x| > ε. – jte Nov 02 '18 at 18:08
  • Your are on the right way. The false definition requires that $|x| = |f(x)-f(0)| < \epsilon$ must be true for all $x$ with $|x| < \delta=2\epsilon$. Maybe you can find one $x$ that fulfills the second condition but not the first? – Ingix Nov 03 '18 at 02:15
  • So can I say that if |x| < ε must be true for all x with |x| < 2ε, there is a clear contratiction as this second condition allows for example |x| = 3ε/2 > ε. – jte Nov 03 '18 at 06:45
  • Exactly. One counterexample is enough. – Ingix Nov 03 '18 at 15:41
  • Alright. Thank you very nuch. – jte Nov 03 '18 at 18:18

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