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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?

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@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

1 Answer 1

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 $_\square?$

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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? –  Thomas E. Feb 28 '13 at 23:15
@Thomas: Yes, I did. –  Brian M. Scott Feb 28 '13 at 23:16

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