I want to evaluate if $f(x)=\frac{1}{x}$ is uniformly continuous.

According to the definition:$\forall\epsilon>0\exists\delta>0\forall(x,x´)\in X\times X\:\:\:d_X(x,x´)<\delta\implies d_y(f(x),f(x´))<\epsilon$

$\delta>0,\:\:|x-c|<\delta$, taking $x=\frac{1}{\delta}$ and $c=x+\delta$. Considering $\epsilon>0$.




Is my proof right?


A few points:

  1. Showing that $a < b > c$ shows does not show any relationship between $a$ and $c$. As such, you can't conclude that $|f(x) - f(c)| > \varepsilon$ from the above argument.
  2. The $x$ and $c$ constructed satisfy $|x - c| = \delta$, not $|x - c| < \delta$. Might I suggest letting $c = x + \frac{\delta}{2}$ instead?
  3. Point 1 can be circumvented by noting that $$\left|\frac{c - x}{xc}\right| = \frac{c - x}{xc} = \frac{\delta}{xc}.$$
  4. I find the best way to disprove uniform continuity is to find a Cauchy sequence in the domain that doesn't map to a Cauchy sequence. For example, $(1/n)_{n=1}^\infty$ is Cauchy, but maps to $(n)_{n=1}^\infty$ which is not Cauchy.

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