What does this notation, $x>x_0(\epsilon)$, mean?

I have seen this in several proofs and haven't quite figured it out.


$x_0(\varepsilon)$ means that $x_0$ depend on $\varepsilon$. For example $$\forall\varepsilon>0, \exists x_0(\varepsilon)>0, \forall x>x_0(\varepsilon), \frac 1{1+x^2}\leq \varepsilon.$$

It's in particular useful when the $x_0$ may depend on several variable; in this case it shows on which variable the $x_0$ depends. An example is the difference between pointwise convergence of a sequence of functions $\{f_n\}$ defined on a set $S$: $$\tag{PC}\forall \varepsilon,\forall x\in S,\exists n_0(\varepsilon,x),\, \forall n\geq n_0(\varepsilon,x),\quad |f_n(x)-f(x)|\leq\varepsilon,$$ and the uniform convergence on $S$: $$\tag{UC}\forall \varepsilon,,\exists n_0(\varepsilon),\, \forall x\in S,\forall n\geq n_0(\varepsilon),\quad |f_n(x)-f(x)|\leq\varepsilon.$$ In these two case, this can be written without marking the dependence, since it's implicit.

  • 6
    $\begingroup$ Formally you could write it as $$\forall\varepsilon>0, \exists x_0>0, \forall x>x_0, \frac 1{1+x^2}\leq \varepsilon$$ but the author wants to remind us that $x_0$ depends on $\varepsilon$. $\endgroup$ – GEdgar Jul 20 '12 at 13:01

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.