3
$\begingroup$

Let $f:[0,1] \cup \{-1\} \to \Bbb R$ be defined by, $f(x)=1 , \forall x \in [0,1]$; $=0$ ,for $x=-1$. Is the function uniformly continuous?

  • here the domain is closed and bounded. Thus compact. And also f(x) is continuous. Thus f should be uniformly continuous. But here how the definition of uniform continuity is applicable? I can't understand it. Or is it uniformly continuous at all?
$\endgroup$
1
  • 2
    $\begingroup$ Since nothing is near $\{-1\}$ then the definition holds trivially. $\endgroup$ May 6, 2017 at 19:01

1 Answer 1

2
$\begingroup$

The "same" reason you saw it is continuous. The function is uniformly continuous on $[0,1]$ clearly. If $x,y \in \{ -1 \}$, then $|x-y| = 0$ and $|f(x)-f(y)| = 0$. So $f|_{\{-1\}}$ is such that for every $\varepsilon > 0$ and every $\delta > 0$ we have $|x-y| < \delta$ implying $|f(x)-f(y)| < \varepsilon$; this is even stronger than the requirement of uniform continuity.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .