Understanding a solution in showing continuity by the inverse image of every open set is open for addition.

Question

Here is the link:

Showing continuity by the inverse image of every open set is open for addition and product function.

But I did not understand (the detailed proof) for the following:

"If $(x_0,y_0)\in\sigma^{-1}(J)$ or equivalently $a<x_0+y_0<b$ then we can find $\epsilon>0$ such that $a<x+y<b$ whenever $|x-x_0|<\epsilon$ and $|y-y_0|<\epsilon$" how can we find this $\epsilon$ is it by Archimedian property? if so how? could anyone explain this for me please?

This was mentioned in the second solution in comments.

Answer 1

It has nothing to do with the Archimedian property.

Take $r>0$ such that $a<x_0+y_0-r<x_0+y_0+r<b$. Now, thake $\varepsilon=\frac r2$. Then, if $\lvert x-x_0\rvert,\lvert x-x_0\rvert<\varepsilon$, then\begin{align}\lvert(x+y)-(x_0+y_0)\rvert&\leqslant\lvert x-x_0\rvert+\lvert y-y_0\rvert\\&<\frac r2+\frac r2\\&=r.\end{align}and therefore$$a<x+y<b.$$

Answer 2

Hint: Take $\epsilon =\frac 1 2\min \{(x_0+y_0)-a,b-(x_0+y_0)\}$. If $|x-x_0| <\epsilon $ and $|y-y_0| <\epsilon $ then $|(x+y)-(x_0+y_0)| <2\epsilon$. Now try to show from this that $x+y \in (a,b)$. [$x+y <x_0+y_0+2\epsilon <b$ etc].