You probably have something else in mind because as it is formulated the answer is yes, any finite subset.
G delta is a countable intersection of open sets; a finite subset means a finite intersection of open sets, which is open.
You look in the wrong direction. A finite set means a set with a finite number of elements. For example a one point set. It is closed and every nonempty G delta set contains one. What is the problem?
Dear Gabriel and George,
Note that in contrast to the a one point set, the emptyset is a better
example for the required closed set.
With best regards,
Arpad Szaz
You can also take an example of any interval [a, b], which is closed and it is an intersection of open intervals (a-1/n, b+1/n), n \in N. This can also be done for infinite closed intervals as well as for countable union of disjoint closed intervals and/or singletons. I think that it is true for any closed subset of the real line R. As any open set in R is a countable union of disjoint open intervals. Any closed set will be a countable union of closed intervals and/or singleton sets. So any closed set in R will be a G_\delta set.
Riddhi Shah, it is not true that every closed set is a countable union of closed intervals or singletons. Take for example the Cantor set.
In the same time, it is true that every closed set in the usual topology of R is G_delta.
Take C a closed set. Then R \ C is a finite or countable infinite union of intervals. I will show only the second case.
Say R \ C = U_k = 1 to infinity of (x_k - r_k, x_k + r_k)
Let F_n = U k = 1 to n [x_k - r_k + r_k/(n + 1), x_k + r_k - r_k/(n+1)]
Then F_n is closed, G_n = R \ F_n is open and
C = intersection G_n
Thank you Gabriel for the correction and the answer to the original question. Just to note to say that the set of G_\delta sets is strictly larger than the set of all closed sets in R as it also contains open closed intervals [a,b) (in fact, it contains any interval). It may be interesting to identify all G_\delta sets. Since any G_\delta set is Borel measurable, the sets which are not Borel or Lebesgue measurable are not G_\delta.
Hi everybody,
A simple proof that every closed set is $G_ \delta$ :
If $F$ is a closed set, then $F= \cap U_n$, where $U_n := \cup_{x \in F } ] x-1/n, x+1/n[.
This simple proof does not make sense to me, can you please elaborate? How do you use that $F$ is closed. Note that any $G_\delta set is Borel measurable.
- Badges
- Science topic
- Similar topics
- Mathematics