A Convention of Set Builder Notation

Question

Given the set $$ X = \{\{a,b\} : a \in \mathbb N ~\wedge~ b \in\{0,1\} \},$$ would the set $\{1\}$ be contained in $X$? I'm not sure how to interpret what happens when $a$ takes on the value $1$ in the set-builder notation. Do we discard $\{1,1\}$ or place it in the set as $\{1\}$?

Answer 1

For $a=b=1$ you have $\{a,b\}=\{1,1\}=\{1\}$, hence $\{1\}\in X$. Recall $$\{a,b\}=\{z:z=a\vee z=b\}$$ thus \begin{align} \{a,a\} &=\{z:z=a\vee z=a\}\\ &=\{z:z=a\}\\ &=\{a\} \end{align} Equivalently, you have $\{a,b\}=\{a\}\cup\{b\}$ so that \begin{align*} \{a,a\} &=\{a\}\cup\{a\}\\ &=\{a\} \end{align*}