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


$\forall\varepsilon>0$ $\exists \delta'>0:\forall x\in A$ $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 $l'=l''$


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$$\lvert x-a\rvert<\min\{\delta_1,\delta_2\}\iff\lvert x-a\rvert<\delta_1\text{ and }\lvert x-a\rvert<\delta_2.$$

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

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