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

I would like to prove the following proposition:

Let $\tau$ be a topology. A finite intersection of elements of $\tau$ is also in $\tau$.

My attempt:

The proof is by induction on the number of elements in the intersection.

Base case: an element of $\tau$ is in $\tau$ by definition.

Suppose that the statement holds for intersections of $k\lt n$ elements. Let $S$ be an intersection of $n$ elements. The intersection of the first $n-1$ elements is in $\tau$ (by the induction hypothesis). Now we have an intersection of two elements of $\tau$ which is an element of $\tau$ (again by the induction hypothesis).

Is my proof correct?

share|cite|improve this question
@Stahl It's an exercise in a book. Is it wrong to call it a theorem? I taught a theorem is just a statement for which there is a proof. – saadtaame Feb 28 '13 at 22:56
@BrianM.Scott how true! everything was perfect up to the unnecessary bracketed remark. Good spotting! – Ittay Weiss Feb 28 '13 at 23:01
@Stahl: The definition that merely requires the intersection of two open sets to be open is at least as common, in my experience. – Brian M. Scott Feb 28 '13 at 23:01
@saadtaame: consider empty intersection too, if it is a finite intersection, – user59671 Feb 28 '13 at 23:03
@saadtaame: No. it yeilds $X$ = whole space. – user59671 Feb 28 '13 at 23:11
up vote 6 down vote accepted

The last step is not correct. The intersection of two open sets is open by definition of topology and not by induction hypothesis.

Consider this:

Theorem: If $A$ is a finite set, all of its elements are equal.

Proof: If $|A|=1$, the claim is trivially true. Suppose the statement holds for finite sets of $k<n$ elements. Let $A$ be a set of $n$ elements. Then the first $n-1$ elements are equal by induction hypothesis. Also the last two elements are equal by induction hypothesis. Therefore all $n$ elements are equal as was to be shown $_\square?$

share|cite|improve this answer
What's wrong with your proof? I can't find the flaw! – saadtaame Feb 28 '13 at 23:01
@saadtaame: The argument used in the induction step doesn’t actually work when $n=1$. – Brian M. Scott Feb 28 '13 at 23:02
@Brian: With Hagen's notations do you mean when $n=2$? Which would be the same as $n=1$ if induction assumption would be for $k\leq n$ instead of $k<n$. Or did I misinterpret something? – T. Eskin Feb 28 '13 at 23:15
@Thomas: Yes, I did. – Brian M. Scott Feb 28 '13 at 23:16

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.