Vtyjkyuk

While solving a problem , I got stuck on the following step: For $A_n = X(omega) leq x_n $ where $x_n$ is an increasing sequence of real numbers converging to say, $x$.

I need to show that union of these $A_n$s is equal to the set $ A= omega in Omega $.

My attempt has been as follows:.

I first showed that the union is contained in this set by taking an element in the union and using that $x_n < x$. But now I'm stuck on proving the other inclusion.

Please help.

Let $omegain A$ or equivalently let $X(omega)<x$.

Then for $n$ large enough we will have $X(omega)leq x_nleq x$ because $x_n$ converges to $x$, so that $xin A_n$.

(Observe that $x_n<X(omega)<x$ for every $n$ implies that $x_n$ does not converge to $x$ because in that case $x-x_n>x-X(omega)>0$ for every $n$)

This shows that: $$Asubseteqbigcup_n=1^inftyA_n$$

Remark: the other inclusion is valid under at least one of the following extra conditions:

• $omegainOmegamid X(omega)=x=varnothing$.


• for every $n$ we have $x_nneq x$ (hence $x_n<x$ in this context).

The second condition will be satisfied if $x_n$ is strictly increasing. Increasing only is not enough.

Let $y<x$. As the sequence converges to $x$ there is an $n$ such that $|x_m-x|<(x-y)$ whenever $mge n$. But then $y<x_m$ for all those $m$.