Let's consider the set [tex]\( D \)[/tex] where [tex]\( D = \{-2, -1, 0, 1, 2\} \)[/tex].
To determine if [tex]\( D \)[/tex] is a subset of itself, we need to understand what it means for one set to be a subset of another. A set [tex]\( A \)[/tex] is a subset of a set [tex]\( B \)[/tex] if every element of [tex]\( A \)[/tex] is also an element of [tex]\( B \)[/tex]. In mathematical terms, [tex]\( A \subseteq B \)[/tex] if for every element [tex]\( x \in A \)[/tex], we have [tex]\( x \in B \)[/tex].
In this particular question, both [tex]\( A \)[/tex] and [tex]\( B \)[/tex] are the same set [tex]\( D \)[/tex]. So we want to check if [tex]\( D \)[/tex] is a subset of itself:
[tex]\[ D \subseteq D \][/tex]
Now let's examine the elements of set [tex]\( D \)[/tex]:
[tex]\[ D = \{-2, -1, 0, 1, 2\} \][/tex]
We need to verify that every element in [tex]\( D \)[/tex] is contained within [tex]\( D \)[/tex]. Clearly, each element [tex]\(-2, -1, 0, 1, 2\)[/tex] is indeed within the set [tex]\( D \)[/tex].
Because every element of [tex]\( D \)[/tex] is contained within [tex]\( D \)[/tex], we can conclude that [tex]\( D \)[/tex] is a subset of itself.
In fact, this is a general property of sets in set theory: a set is always a subset of itself. This is because the definition of a subset is inherently satisfied when considering the set with itself.
To summarize:
- Require: [tex]\( D \subseteq D \)[/tex]
- Check: Every element in [tex]\( D = \{-2, -1, 0, 1, 2\} \)[/tex] is present in [tex]\( D \)[/tex]
- Conclude: [tex]\( D \)[/tex] is indeed a subset of itself
Therefore, [tex]\( D \)[/tex] is a subset of itself, as every element of [tex]\( D \)[/tex] is contained within [tex]\( D \)[/tex].
Final answer:
Yes, [tex]\( D \)[/tex] is a subset of itself because every element of the set is contained within the set.
