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Vacuous truth in regards to vector spaces

  1. Nov 13, 2008 #1
    The only reading on vacuous truth has been from Wikipedia, so I may be misunderstanding something here. Anyway, I was skimming through a Linear Algebra textbook and it said that the empty set is NOT a subspace of every vector space. But I was thinking, shouldn't this be vacuously true? For example, one of the conditions that a set must meet in order for it to be a vector space is that given two elements of the set x and y, x + y = y + x. Isn't this vacuously true for the empty set because there are no elements in the set at all?
     
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  3. Nov 13, 2008 #2

    nicksauce

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    My thought would be that the empty set does not satisfy the requirement of a zero vector, but I might be wrong.
     
  4. Nov 13, 2008 #3
    I was thinking about it some more, and I think what I said about it being vacuously true is false now. Similar to what you said, suppose V is a vector space over a field F. Then there must be an element, 1, which belongs to F, such that 1*x = x for every x in V. But of course there isn't a 1 in the empty set, so this condition isn't satisfied.

    And just what you were saying, there has to be a zero vector such that 0 + x = x for some x in V. But of course there's no 0 in the empty set.

    Is this reasoning right?
     
  5. Nov 14, 2008 #4

    HallsofIvy

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    No, your reasoning is not right. It was the subspace that you wanted to be empty not the field! Part of the definition that a subset be a subspace is that it contain the 0 vector which means it cannot be empty.

    Recently there was another thread in which a person gave the as the requirements that subset of a vector space be a subspace:
    1. The 0 vector is in the subset
    2. The set is closed under additon
    3. The set is closed under scalar multiplication
    4. If x in the set, -x is also in the set

    and it was argued that only (2) and (3) are necessary since, if x is in the set, by (3) (-1)x= -x is in the set and then, by (2), x+ (-x)= 0 is in the set. That is not correct since it fails if there is NO x in the set: By just (2) and (3), the empty set would be a subspace of any vector space. Of course, (1) and (4) could be replace by the simple requirement that the set be non-empty.
     
  6. Nov 14, 2008 #5
    What turns out to be true vacuously is largely a matter of definition. We choose our definitions carefully, though, so we don't have to disclaim stupid, trivial counter-cases in our theorems.

    In this particular case, you could simply work with a definition of a vector space removing the "contains the zero vector" property (since it's implied by the closure under scalar multiplication anyway), and then yes, {} would be a vector space.

    But the point of mathematics isn't usually to work out that's trivial =-)
     
  7. Nov 14, 2008 #6
    Since the dimension of {0} is 0, what would the dimension of the empty set be if it was accepted as a vector space(subspace)? Nonexistent. What would be a basis for it? Nonexistent. Seems like a rather useless definition.
     
  8. Nov 14, 2008 #7
    Like I said, you could do it, but it would require putting annoying qualifications on theorems and definitions.

    Dimension can mean whatever you want, but the most natural way to extend it with this definition would be that only non-empty spaces can have dimension.
     
  9. Nov 14, 2008 #8
    Halls, I understand what you're saying. I guess the simplest way to see that the empty set isn't a subspace is to see that it doesn't contain the zero vector. But reading this part from Wikipedia: "An even simpler example concerns the theorem that says that for any set X, the empty set is a subset of X. This is equivalent to asserting that every element of the empty set is an element of X, which is vacuously true since there are no elements of the empty set"

    Applying this to the empty set being a subspace of V, can't you say that x + (-x) = 0 since there is no x in the empty set?
     
  10. Nov 14, 2008 #9

    Office_Shredder

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    for all x in the empty set, x+(-x)=0 (proof by lack of counterexample). We also know that if x is in the empty set, then -x is (again trivially true, there are no x in the empty set , and for all x,y in the empty set, x+y is in the empty set. So you almost think that x+(-x)=0 is in the empty set, except that you had no x in the empty set to start with, so there's no x to add to its negative to begin with
     
  11. Nov 14, 2008 #10
    After doing more reading about statements being vacuously true, I think I fully understand now why the empty set is not a subspace of every vector space. The statement "If P then Q" is false only if P is true and Q is false. If P is false and Q is true, the statement is still true. If P is false and Q is false, the statement is still true. So with the statement "If a set is empty, it is a subspace of every vector space", P is true as we can obviously find an empty set, but Q is false because for one reason, it does not contain the zero vector. It also doesn't contain the vector "1" such that x*1 = x.
     
    Last edited: Nov 14, 2008
  12. Nov 14, 2008 #11

    Office_Shredder

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    JG, there is no "1" vector. There's a "1" in the field the vector space is over, but since the empty set is over the same field as your original vector space, you still have that element in the field the empty set is defined over
     
  13. Nov 14, 2008 #12
    You're right. I got too sloppy with my terms.
     
  14. Nov 14, 2008 #13

    HallsofIvy

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    To be "vacuously true", there must be an hypothesis "if ... " that is false. Since x+(-x)= 0 is true for every member of the original vector space, I don't see any point in talking about the zero vector. Saying that "x+ (-x)= 0" or even "if x is in the empty set, then x+ (-x)= 0" or, for that matter, the "vacuously true" statement " if x is in the empty set, then 2x= 0", saying nothing about 0 actually being in the empty set.

    Perhaps you are thinking that, since it is always true that "x+(-x)= 0" we can say "if x is in the empty set then x+ (-x) is in the empty set" is vacuously true. That is correct but says nothing about whether "x+ (-x) is in the empty set". When we say "if A then B" is "vacuously true" if A is false, we are saying the entire "if... then ..." statement is true whether or not the conclusion is true.
     
  15. Nov 14, 2008 #14
    Halls, when you say "When we say "if A then B" is "vacuously true" if A is false, we are saying the entire "if... then ..." statement is true whether or not the conclusion is true.", is this because If A is false and B is true, the whole "if..then.." statement is true, and If A is false and B happens to be false as well, the "if..then..." statement is also true, so we cannot say whether B is true or not?

    If this is so, then I see how we cannot conclude that x+(-x) = 0 is in the empty set.

    I have two more questions:

    1) Is what I said in my other post correct: "The statement "If P then Q" is false only if P is true and Q is false. If P is false and Q is true, the statement is still true. If P is false and Q is false, the statement is still true. So with the statement "If a set is empty, it is a subspace of every vector space", P is true as we can obviously find an empty set, but Q is false because for one reason, it does not contain the zero vector."


    2) How come if P is false and Q is false, we can say that the statement "If P then Q" is true? I was trying to reason it out and the furthest I got was that if Q is false when P is false, then Q must bet true when P is true. But I think this is wrong because just because Q is false and P is false doesn't mean that Q is false whenever is false.
     
  16. Nov 15, 2008 #15

    HallsofIvy

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    Yes. that is correct.

    Yes, in that case, P is true and Q is false so thestatement "If P is an empty set is a subspace of every vector space" is false.


    [/quote]2) How come if P is false and Q is false, we can say that the statement "If P then Q" is true? I was trying to reason it out and the furthest I got was that if Q is false when P is false, then Q must bet true when P is true. But I think this is wrong because just because Q is false and P is false doesn't mean that Q is false whenever is false.[/QUOTE]
    I don't know what you mean by "whenever P is false". A statement is either true or false. There is no "whenever" about it. And the statement "if P then Q" does not say anything about what happens if P is false. If P is false and Q is false then "If P then Q" is a true statement. If P is false and Q is true the "If P then Q" is still a true statement.

    In any case, "If P then Q" being a true statement whenever P is false regardless of whether Q is true or false is a definition. But it does match "everyday" usage. If a teacher says "If you get an A on every test, you will receive an A as a final grade", he is not saying anything at all about what will happen if you do NOT get an A on every test. If you get B's on every test and do not get an A, was the statement false? Of course not. If you get A's on everytest but one and a B on that and get an A for the course, was the statement false? Of course not.

    I think of it as "innocent until proven guilty".
     
  17. Nov 15, 2008 #16
    Thanks for the analogy. I understand now :)
     
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