Countable union in Borel sigma algebra

If Borel sigma algebra is defined to be the sigma algebra generated by the class of all open intervals of $\Omega$=(0, 1], then all these open intervals will be uncountable. My question is how this sigma algebra will be closed under the formation of countable unions (or intersections) while I can't count its elements ? how would I take countable unions of uncountable elements.

• Choose countably many elements of the algebra, and take the union of those. Then regardless of which elements were chosen, the result has to still be an element of the algebra. – Ben Sheller May 7 '16 at 20:29
• but in this way there will still be some elements existing in the sigma algebra but their union is not there, doesn't that contradicts with the condition that it must be closed under unions ? – Dorgham May 7 '16 at 20:34
• No, any countable union of elements in the sigma algebra is in the sigma algebra. So if you want a particular set $X$ in the sigma algebra to be involved in a countable union, just take $X$ and countably many other sets in the sigma algebra (it doesn't matter what they are), and take the union of $X$ together with the union of those sets. The result must lie in the sigma algebra. – Ben Sheller May 7 '16 at 20:36
• Real numbers aren't members of the sigma algebra; sets of reals are. – BrianO May 7 '16 at 20:52
• Yes, the number of these sets is uncountable: it's at least $2^{\aleph_0}$, as it contains all open subintervals, and there are continuum-many of those. The sigma algebra is the larger collection you get by applying $\setminus$ and countable $\bigcup, \bigcap$, countably many times to open sets. – BrianO May 7 '16 at 22:38