0
$\begingroup$

I was doing both proof of uniqueness of limit of a sequence and function and i don't understand why in sequence we take $\nu=max\{\nu',\nu''\}$ and in function we take $\delta=min\{\delta',\delta''\}$

I hope it's clear without writing the whole proof of both theorems, thanks.

Let X,Y be metric spaces, $f:A\to Y$ a function and $l',l''\in Y$

if $\lim_{x\to x_0}f(x)=l'$ and $\lim_{x\to x_0}f(x)=l'$ $\implies l'=l''$

Proof.

$\forall\varepsilon>0$ $\exists \delta'>0:\forall x\in A$ $d_X(x,x_0)<\delta $ $x\neq x_0$ $d_Y(f(x),l')<\varepsilon$

$\forall\varepsilon>0$ $\exists \delta''>0:\forall x\in A$ $d_X(x,x_0)<\delta $ $x\neq x_0$ $d_Y(f(x),l'')<\varepsilon$

Let $\delta=min\{\delta',\delta''\}$. So

$\forall\varepsilon>0$ $\exists \delta>0:\forall x\in A$ $d_X(x,x_0)<\delta $ $x\neq x_0$ $d_Y(f(x),l')<\varepsilon$ $d_Y(f(x),l'')<\varepsilon$

Then $\forall>0$ $d_Y(l',l'')<2\varepsilon$ then $l'=l''$

$\endgroup$
2
$\begingroup$

Yes, it is clear. When dealing with sequences, we have a condition of the type$$(\exists\nu\in\mathbb{N}):n\geqslant\nu\implies C_n$$So, if we have a $\nu_1$ and a $\nu_2$ such that $n\implies C_n$ whenever $n\geqslant\nu_1$ and $n\geqslant\nu_2$, then we take $\nu=\max\{\nu_1,\nu_2\}$ because$$n\geqslant\max\{\nu_1,\nu_2\}\iff n\geqslant\nu_1\text{ and }n\geqslant\nu_2.$$But in the case of functions, we have a condition of the type$$\lvert x-a\rvert<\delta\implies\cdots$$Before we had $\geqslant$ and now we have $<$. So, we consider $\min\{\delta_1,\delta_2\}$, because$$\lvert x-a\rvert<\min\{\delta_1,\delta_2\}\iff\lvert x-a\rvert<\delta_1\text{ and }\lvert x-a\rvert<\delta_2.$$

$\endgroup$
  • $\begingroup$ Oh thanks, it's pretty obvious now. $\endgroup$ – Archimedess Dec 1 '18 at 19:53

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.