(a) You didn't use the specific definitions of the sets. This is not a generic question about any three sets.
(b) Your explanation is missing something.
(c) Okay
(d) You're probably right, but if I were required to base my answer only on what we are explicitly told, I might say "insufficient data". I could say the same about (b) and (c), since we aren't told that "aardvark" is in the dictionary ...
(e, f) I think you are supposed to explain in terms of the dictionary and the definitions you were given: "words that ..."
(g) Your reasoning is not stated fully, but you are right.
In questions a)b)c)d) We were only supposed to state whether statement is true or false, explanations are there on my part only.
a) I explained why i think this statement is false (because of definitions of the sets..)
As pka pointed out, A∪B consists of all words that are either before dog or after cat, which is the entire dictionary. If a word is not before dog, it is after cat. So any subset of the dictionary, such as C, is a subset of A∪B.Union of A and B are all the words that appear before the dog, and all the words that appear after the cat. C are all the words that have two identical consecutive letters.
Question: is a C subset of union of A and B?
A set is a subset of a set if all of its members are also members of superset. Because not all words that have two identical consecutive letters also go before dog and after cat, so that statement is false. I could be wrong, but im pretty sure its false.
You have told twice that \(A\cup B\) is the set of all words in the dictionary.Union of A and B are all the words that appear before the dog, and all the words that appear after the cat. C are all the words that have two identical consecutive letters. Question: is a C subset of union of A and B?
A set is a subset of a set if all of its members are also members of superset. Because not all words that have two identical consecutive letters also go before dog and after cat, so that statement is false. I could be wrong, but im pretty sure its false.