# Misunderstanding with the union of open sets

In the axioms that define open sets, there is the condition says that any union of open sets is an open set. In some textbooks this condition is translated in simbols as follows:

For every family of open sets $\{V_i\}_{i\in I}$ then $\bigcup_i V_i$ is open.

But formally, since $I$ is a set of indices it must have countable cardinality, so in this way it seems that only a countable union of open sets is open.

The same misunderstanding is present for coverings of a set, so when one says that $V$ admits an open covering it isn't clear if it is made with a countable or uncountable number of sets.

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Are you the same user as this one? If so, you can contact a moderator to have your accounts merged. –  Zhen Lin May 24 '12 at 11:30
Yes I'm the same, I had some problems with the registration. Thanks for the advice. –  Dubious May 24 '12 at 11:33

Any set $I$ can be an index set. So an index set can have any cardinality, however large (or small).
To elaborate a bit on this, consider the following family of open sets of $\Bbb R^2$: For each element $x\in\Bbb R^2$, let $V_x$ be the open disc of radius 1 centered at $x$. The index set $I$ here consists of $\Bbb R^2$, which is uncountable. –  MJD May 24 '12 at 11:31
Another way to see that the set $I$ needn't be countable is this. Let $I$, itself, be any collection of open sets (countable or otherwise). Then for $i\in I$, we define $V_i:= i$, giving us the desired indexing, and letting us write the union of the $I$-sets as the indexed union $\bigcup_{i\in I}V_i$. We can do this in many circumstances. The upshot is that we can always index a set by its own elements, if we like.