Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 12 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.