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I don't understand the Heine-Cantor theorem because of one example: The function $x\to \frac{1}{x}$ is not uniform continuous, and we can clearly see in the graph just by looking at the interval $[1,10]$ for instance. The Heine-Cantor theorem doesn't seem to work for $x\to \frac{1}{x}$ but I know that I'm wrong.

  • Clearly from the graph... – Karl Kroningfeld Jun 05 '15 at 00:06
  • i still don't get it, why does the theorem work for x->1/x –  Jun 05 '15 at 00:08
  • Since the function $x\to\frac1x$ is uniformly continuous on the interval $[1,10],$ it's apparent that you don';t know what "uniformly continuous" means. What is the definition of uniform continuity? – bof Jun 05 '15 at 00:22
  • The teacher had explained it today in the classroom but he didn't show us any graphic example, so i don't think i've really understood the concept. From what i've understood, if f is uniformly continuous, then if you take an epsilon, there exists an alpha such that, for every x,y, if the distance between x and y is lower than alpha, then the distance between their images is lower than epsilon –  Jun 05 '15 at 00:26
  • my problem is that i can't identify on which set f is uniformly continuous –  Jun 05 '15 at 00:27

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$f(x)=1/x$ is uniformly continuous on $[1,10]$, so I'm not sure what you clearly see from the graph. The function $f(x)=1/x$ is not uniformly continuous on $(0,1]$, but that is fine since the Heine-Cantor theorem states that a continuous function on a closed and bounded interval is uniformly continuous. But $(0,1]$ is not closed.

Ittay Weiss
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