Suppose S is the set of all things that are blue or green. Then my mug is in S because it's green and therefore satisfies "x is blue or x is green," and my pen is in the set S because it's blue and therefore satisfies "x is blue or x is green." Now it's true: satisfying "x is blue or x is green" picks out only one set: the set of all things that satisfy "x is blue or x is green." But the condition "x is green" is a different condition, and so is "x is blue."

However: when you say "being blue or green cannot be the reason why any other object is in any other set," there's an ambiguity. That could be read as "being blue cannot be the reason why an object is in any other set and being green cannot be the reason why an object is in any other set." In that case, however, it's false. Being green, and hence satisfying "x is green" puts my mug in the set G of all green things, and in the set S of all things that are either green or blue—that satisfy "x is green or x is blue." These two sets are not the same. One is a proper subset of the other. Being in the set S doesn't entail being in the set G, and also doesn't entail being in the set B, though it does entail being in either G or B.

The key is to formulate the membership condition so that there's no room for ambiguity. We have

o is in S if and only if o satisfies "x is green or x is blue."
o is in G if and only if o satisfies "x is green"
o is in B if and only if o satisfies "x is blue"

These are three different conditions that pick out three different sets. G and B are disjoint from one another and are proper subsets of S. The union of G and B is S. But the formulation "x is green or blue," while not wrong, masks the fact that "x is green or blue" amounts to "x is green or x is blue," and so is the disjunction (the "or") of two conditions. An object is in S if it satisfies either of those conditions. It's in G only if it satisfies the first, and in B only if it satisfies the second. But clearly by virtue of satisfying one condition ("x is green") my mug can be in S and also in G.

## Suppose S is the set of all

Suppose S is the set of all things that are blue or green. Then my mug is in S because it's green and therefore satisfies "x is blue or x is green," and my pen is in the set S because it's blue and therefore satisfies "x is blue or x is green." Now it's true: satisfying "x is blue or x is green" picks out only one set: the set of all things that satisfy "x is blue or x is green." But the condition "x is green" is a different condition, and so is "x is blue."

However: when you say "being blue or green cannot be the reason why any other object is in any other set," there's an ambiguity. That could be read as "being blue cannot be the reason why an object is in any other set and being green cannot be the reason why an object is in any other set." In that case, however, it's false. Being green, and hence satisfying "x is green" puts my mug in the set G of all green things,

andin the set S of all things that are either green or blue—that satisfy "x is green or x is blue." These two sets are not the same. One is a proper subset of the other. Being in the set S doesn't entail being in the set G, and also doesn't entail being in the set B, though it does entail being in either G or B.The key is to formulate the membership condition so that there's no room for ambiguity. We have

o is in S if and only if o satisfies "x is green or x is blue."

o is in G if and only if o satisfies "x is green"

o is in B if and only if o satisfies "x is blue"

These are three different conditions that pick out three different sets. G and B are disjoint from one another and are proper subsets of S. The union of G and B is S. But the formulation "x is green or blue," while not wrong, masks the fact that "x is green or blue" amounts to "x is green or x is blue," and so is the disjunction (the "or") of two conditions. An object is in S if it satisfies either of those conditions. It's in G only if it satisfies the first, and in B only if it satisfies the second. But clearly by virtue of satisfying one condition ("x is green") my mug can be in S and also in G.