by Corry Shores
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[The following is summary. My commentary is in brackets. Boldface is mine.]
Summary of
Patrick Suppes
Introduction to Logic
Ch. 9 Sets
§9.4 The Empty Set
Brief summary:
The empty set has no members. It is a subset of all other sets, even if it is not explicitly specified as a member. And the empty set is the only set that can be a subset of the empty set.
Summary
[In section 9.2 we mentioned the empty set, symbolized Λ.] We earlier noted that the empty set has the following property: for every x, x∉ Λ.
We now add that although the empty set has no members, it itself can still be a member of other sets.
Thus if we speak of the set of all subsets of the set {1, 2}, we are speaking of the set {{1, 2}, {1}, {2}, Λ} which has four members; the three-member set {{1, 2}, {1}, {2}} on the other hand, is the set of all non-empty subsets of {1, 2}.
(184)
[We now get the reasoning for why the empty set is a member of every other set. The reasoning seems to depend on the truth-functionality of conditionals. But for that reason, it is hard for me to extract the intuition. The way we come to this conclusion will be by denying the antecedent in a conditional, which will make it true. The conditional is in the definition for subset inclusion, namely, A⊆B↔(x)(x∈A→x∈B). So look first at (x∈A→x∈B). Suppose we were to show that x∈A is false. That makes (x∈A→x∈B) true, and thus that makes A⊆B true. We will now consider A as being the empty set Λ. We defined it as (x)(x∉Λ). This makes the antecedent false, thus the conditional true, thus the left side of the biconditional true, thus Λ⊆B, or in other words, the empty set is a subset of every other set. That much is easy to follow. But how do we understand the insight in the reasoning here and not just the mechanics of the proof? Something qualifies as a set’s subset if all that subset’s members are also members of the first set. Perhaps the idea is that if we stipulated that sets do not contain the empty set as a subset, we are implying that the empty set has members which the other sets do not have. But this is false, since it does have any members. This forces us to consider the empty set as a subset of all other sets. But still I am not entirely sure I have the intuition of this reasoning. Let me quote:]
We recall the fact that a set A is a subset of a set B if and only if every member of A is also a member of B, i.e., if and only if: for every x, if x ∈ A, then x ∈ B. In particular, the empty set Λ is a subset of a set B if and only if: for every x, if x ∈ Λ, then x ∈ B. Since always x ∉ Λ, however, it is always true that if x ∈ Λ, then x ∈ B. Thus, for every set B, we have:
Λ ⊆ B.
That is, the empty set is a subset of every set.
(184)
[It also follows that the empty set is the only set included in the empty set. I am not sure why we do not instead say that not even the empty set is included in the empty set. (Perhaps it is for similar reasoning, namely, that the empty set does not have any members that the empty set lacks, and therefore it must be a subset). At any rate, we can at least be sure that no other set is a member of the empty set. The reasoning here is that we already established that the empty set is a subset of every other set. If we say that one of these other sets is also a subset of the empty set, then we are fulfilling the definition for the equality of sets. But surely not every non-empty set is identical with the empty set. Thus it cannot be that any other set is found in the empty set. I quote:]
In addition, the empty set is the only set which is a subset of the empty set; for if B ⊆ Λ, then, since we also have: Λ ⊆ B, we can conclude that B = Λ.
(184)
Suppes then summarizes these facts about the empty set in the following formulas:
(i) (x)(x ∉ Λ),
(ii) (∃A)(Λ ∈ A) & (∃A)(Λ ∉ A),
(iii) (A)(Λ ⊆ A),
(iv) (A)(A ⊆ Λ ↔ A = Λ).
(Suppes 184)
[The first one, (x)(x∉Λ), “for all x, x is not a member of the empty set lambda”, we know from before. It defines the empty set as the one for which no members belong. The second one, (∃A)(Λ∈A)&(∃A)(Λ∉A), I am less sure about, because it almost seems like a contradiction. It could perhaps be read, “for some set A, the empty set is a member of A, and for some set A, the empty set is not a member of A”. I am not sure how to be clear on why this is not a contradiction and also what insight it is expressing. My best guess is that it means to convey the idea in the part of the text where Suppes writes: “if we speak of the set of all subsets of the set {1,2}, we are speaking of the set {{1,2},{1},{2},Λ} which has four members; the three-member set {{1,2},{1},{2}} on the other hand, is the set of all non-empty subsets of {1,2}” (184). So here we have set {1,2}, which may be an illustration for the A in the second formulation above. We then speak of {1,2} in two ways, namely, in terms of all its subsets and in terms of all its non-empty subsets. Λ is a subset for the first sense but not for the second. It is also possible that we are not to interpret A as the same set in both parts of the conjunction. But then I wonder in that case why we would not use another letter, and also, I do not know what insight it would be expressing. The third formulation ((A)(Λ⊆A)), makes the point that the empty set is a subset of all other sets. This again might seem to present us with a contradiction, because we said above that there is a set for which the empty set is not a member. However, we might need to distinguish membership with inclusion. A set is included as a subset of another set if all the first one’s members are also members of the second set. This we said means that the empty set is always a subset of other sets. However, that does not mean the empty set is a member of all other sets. To be a member set is different than being a subset. Somehow, it is possible for a set to have the empty set as a subset but not as a member. I am not exactly sure how. But let us look again at the set: {1,2}. It has two members: 1, and 2. Now we want to form a new set, which is the set of all the subsets of {1,2}. I suppose this is like asking, what are the different ways this set can be internally grouped? One grouping is all the members, {1,2}. Another group is a group with just one member, {1}. A third group is another group with just one member, {2}. And a final group is a group with no members Λ. As we said, we were constructing a set of all these subsets, so we put them all within another pair of braces: {{1,2},{1},{2},Λ}. Now, we also said that we can construct a set of all of {1,2}’s subsets, only this time we stipulate that we just want all the non-empty subsets. This gave us {{1,2},{1},{2}}. Now, recall the formulation we are trying to explain, (A)(Λ⊆A), which seemed to potentially contradict (∃A)(Λ∉A). In our example, we might for instance ask, why is Λ excluded from {{1,2},{1},{2}}, if it is always a subset of other sets? Perhaps the idea here is that it is a subset of {{1,2},{1},{2}}, but it just not a member. In other words, we can regroup {{1,2},{1},{2}} into another set where Λ is found, that being the set of all subsets in {{1,2},{1},{2}}. But that does not mean the empty set was originally a member. It just means that the empty set is one of the groupings that would result from finding all its possible internal regroupings. I am not sure. Another thing to note is that if we were to find all the subsets of {{1,2},{1},{2}}, perhaps we would have more than just {{1,2},{1},{2},Λ}. Perhaps we would have {{1,2},{1},{2},Λ,{{1,2},{1}},{{1,2},{2}}}. And in that case, perhaps the idea is that whenever we find all the subsets of a set, we are always creating a new larger set, so we should not be surprised that the new larger set has members not listed in the original, like the empty set for example. And therefore, the fact that both (A)(Λ⊆A) and (∃A)(Λ∉A) hold should not seem like a contradiction, because Λ need not be a member of some set in order for it to be found in the set of all that set’s subsets. The fourth formula ((A)(A⊆Λ↔A=Λ)) says that the only subset that can be included in the empty set is the empty set itself.]
Suppes, Patrick. Introduction to Logic. New York: Van Nostrand Reinhold / Litton Educational, 1957.
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