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

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

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$x_0(\varepsilon)$ 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.

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    $\begingroup$ 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

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