3
$\begingroup$

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.

  • 0
    Yes I'm the same, I had some problems with the registration. Thanks for the advice.2012-05-24

1 Answers 1

12

Any set $I$ can be an index set. So an index set can have any cardinality, however large (or small).

  • 7
    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.2012-05-24