- #1
evinda
Gold Member
MHB
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Hi! (Smile)
I want to prove that for each set $A$:
$$A \subset \mathcal P \cup A$$
According to my notes, we prove it like that:
Let $x \in A$. We want to show that $x \in \mathcal P \cup A$, so, that: $\exists y \in \mathcal P \cup A$, such that $x=y$.
It suffices to show that if $z \in x$, then $z \in \cup A$, that stands, since $x \in A$.
Could you explain me why we want to show that :
$$\exists y \in \mathcal P \cup A, \text{ so that } x=y$$
?
At the beginning, we suppose that $x \in A$ and we want to prove that $x \in \mathcal P \cup A$, so we want to prove that $x \subset \cup A$.
I haven't understood what we could do to prove this.. (Worried)
I want to prove that for each set $A$:
$$A \subset \mathcal P \cup A$$
According to my notes, we prove it like that:
Let $x \in A$. We want to show that $x \in \mathcal P \cup A$, so, that: $\exists y \in \mathcal P \cup A$, such that $x=y$.
It suffices to show that if $z \in x$, then $z \in \cup A$, that stands, since $x \in A$.
Could you explain me why we want to show that :
$$\exists y \in \mathcal P \cup A, \text{ so that } x=y$$
?
At the beginning, we suppose that $x \in A$ and we want to prove that $x \in \mathcal P \cup A$, so we want to prove that $x \subset \cup A$.
I haven't understood what we could do to prove this.. (Worried)