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Today in my math class I presented a counter example to the theorem: "if any infinite sequence in X has an adherent point in X, then X is compact."

Let $X=(-1,2)$. Choose $\{X_{n}\}= \frac{1}{n} = \{1,\frac{1}{2}, \frac{1}{3},\cdots \}$. Then $0$ is an adherent point of $\{ X_{n} \}$ in $X$. But X is not compact.

My professor told me that "if any" is synonymous with "if every" in this instance so my counter example doesn't work but I can't see how that is the case. Can anyone give me some insight into how these statements are equivalent? Sorry about any poor formatting I am on mobile.

Leucippus
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Andrew Holman
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  • https://en.wikipedia.org/wiki/Sequentially_compact_space#Examples_and_properties – Will Jagy Nov 20 '15 at 21:32
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    Your professor is right, there isn't a distinction between "if any" and "if every" in mathematics. You might be reading the text as "if there exists any sequence..." rather than "if, for any sequence..." – Thomas Andrews Nov 20 '15 at 21:33
  • This is a good book, counterexamples in this topic are sometimes difficult, you would not expect to find some of them yourself: https://en.wikipedia.org/wiki/Counterexamples_in_Topology – Will Jagy Nov 20 '15 at 21:35
  • Inexpensive: http://store.doverpublications.com/048668735x.html – Will Jagy Nov 20 '15 at 21:38
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    If any means there exists such a sequence. If every means all sequences. – Yunus Syed Nov 20 '15 at 21:38
  • Thomas, thank you that small distinction makes sense to me now. I was originally interpreting it as "if there exists any". – Andrew Holman Nov 20 '15 at 21:41
  • @WillJagy: That is an excellent book, but it is not addressing this question. OP has found a counterexample to his reading of the theorem and is asking why his reading of the theorem is not correct. – Ross Millikan Nov 20 '15 at 21:41
  • @YunusSyed No, you are incorrect. "If any" is probably best avoided because of the colloquial usage, but it doesn't mean "if exists." – Thomas Andrews Nov 20 '15 at 21:47
  • @Ross, I know. People with more patience than I have gave excellent comments and answers, and the OP seems satisfied now. For me, half the battle in dealing with anything is examples, as in the famous story about Hassler Whitney in the preface to the book Conics by Keith Kendig. – Will Jagy Nov 20 '15 at 21:47
  • @Ross, https://books.google.com/books?id=TeWCKNsSy6wC&pg=PR12&lpg=PR12&dq=conics+kendig+whitney&source=bl&ots=jksT2RRRa6&sig=mwseEelgPxzfLNiq6FlHoEmsXbM&hl=en&sa=X&ved=0ahUKEwjJ2oPM_Z_JAhXUNIgKHeSCDCIQ6AEIHjAA#v=onepage&q=conics%20kendig%20whitney&f=false – Will Jagy Nov 20 '15 at 21:50
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    I agree that "if any" is probably best avoided. That statement of the theorem is ambiguous. It wouldn't be if it said "if every infinite sequence...". – BrianO Nov 20 '15 at 22:16
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    I would go further than BrianO: "If any" intended to mean "if every" is an indefensible misuse of English. If any mathematician disagrees with me, then (according to that mathematician) every mathematician does! But Thomas Andrews on the one side taken with BrianO and Yurus Syed on the other provide a counterexample to this. – Rob Arthan Nov 21 '15 at 00:21
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    @Thomas Andrews: suppose I say that a set of vectors is linearly dependent if any finite subset of it sums to zero. That is a true theorem, written in normal idiomatic mathematical English, and the "if any" there certainly does not mean "if every". Similarly "If a subset of a vector space is nonempty, closed under vector addition, and closed under scalar multiplication, it is a subspace. If any of those three conditions fails, the subset is not a subspace." – Carl Mummert Nov 23 '15 at 12:50
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    @Rob Arthan: having answered this question several times now on Math.SE, I think the number of people who instinctively guess "any" means "for all" is at least as large as the number who instinctively guess "any" means "for some". Actually, examples of both usages are easy to find among professional mathematicians. – Carl Mummert Nov 23 '15 at 22:06
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    @CarlMummert: I agree. So we should all write carefully to avoid falling into these potential tarpits of ambiguity. – Rob Arthan Nov 23 '15 at 22:26

1 Answers1

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If you are in doubt, sometimes, putting parentheses around logical statements can help make their meaning clearer. In this instance:

if (any infinite sequence in $X$ has an adherent point in $X$) then ($X$ is compact)

What the first bracketed statement essentially means is "choose any infinite sequence in $X$ and it will have an adherent point in $X$". Now, at this point, it should be quite clear that replacing 'any' with 'every' would give you an equivalent statement.

ryang
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L. T. P. L.
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    Thanks. This is a good trick and definitely helps to clarify my understanding of the statement. – Andrew Holman Nov 20 '15 at 21:44
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    Yes, but that's not where the brackets go in English: "if any" acts like a logical quantifier and separating the "if" and the "any" doesn't make linguistic sense. You can't usefully redefine a natural language by pretending it conforms to logical and structural laws of your own invention. – Rob Arthan Nov 21 '15 at 00:28
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    I suggest reminding yourself of the formal definition of 'any'. The cause of this debate is that any is misused so frequently in colloquial english. – L. T. P. L. Nov 21 '15 at 12:52
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    @L.T.P.L.: like every other natural language, English does not have a formal definition: its meaning is defined by the collective agreement of its speakers, listeners, readers and writers. Natural languages are not governed by logical laws and you should not dismiss standard English idioms as "misuse" just because they don't fit into a logical framework of your own devising. – Rob Arthan Nov 22 '15 at 00:28
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    It's not "my own devising", 'any' means whichever items of a specified class you choose.

    For example, "if any infinite sequence in X has an adherent point in X then..." means "if whichever infinite sequence in X you choose has an adherent point in X then..."

    Please stop arguing - the sooner you can accept that you fundamentally misunderstand this, the sooner you can learn, move on, and stop misinforming people.

     – L. T. P. L. Nov 22 '15 at 11:58
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    But I don't misunderstand this. You are expressing an opinion about the English language that is not justified by standard usage. The only sensible advice for mathematicians writing in English is to avoid these tricky issues. I am not misinforming anybody, but you are (by claiming that words like "any" can be interpreted without taking into account their context). – Rob Arthan Nov 23 '15 at 00:11
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    @L.T.P.L.: PS. Your latest example is particularly uncompelling: "if whichever infinite sequence in $X$ you choose has an adherent point in $X$ then ..." isn't good English. "If any infinite sequence in $X$ has an adherent point in $X$ then ..." is good English, and means "If there exists an infinite sequence in $X$ that has an adherent point in $X$ then ...". The English language is being astonishingly logical here in treating $A \Rightarrow B$ to mean $\lnot A \lor B$ and then correctly negating the universal quantifier. – Rob Arthan Nov 23 '15 at 00:23
  • I disagree with this answer; my response here is consistent with what Rob Arthan says. – ryang Mar 10 '23 at 13:22