[Math] Confusion about the null (empty) set being contained in other sets

elementary-set-theory

I'm having a tough time understanding how the set theory of null sets work. I have:
$$
X=\emptyset,\quad\quad Y = \{\emptyset\},\quad\quad Z = \{\{\emptyset\}\}.
$$

Some of my self-study exercises include these true or false questions. Now, I'm more concerned with the reasoning behind why they're true or false as opposed to the answers as I already have the answers, I just want the understanding.

  1. $\emptyset \in X$.

    I know this is false because the null set is not an element of any set.

  2. $\emptyset \in Y$.

    I don't know why this is true.

  3. $\emptyset \in Z$.

    I don't know why this is false.

  4. $X \subseteq Y$.

    I know this is true because the null set symbol is directly within the set.

  5. $Y \subseteq Z$.

    I don't know why this is true.

  6. $X \in Y$.

    The same reason why (2) is true, I understand this one.

  7. $Y \in Z$.

    This is true because $\{\emptyset\}$ is directly within the set defined by $Z$.

Best Answer

Very important facts: Set membership ($\in$) and set inclusion ($\subseteq$) are two very different things. When we write $A = \{b\}$, it implies that $b \in A$ but it does not imply that $b \subseteq A$. The only way for $b \subseteq A$ to be true is if $b$ is a set and the set $b$ does not contain any members that are not also members of $A$.

Now to apply these facts:

  1. The null set can be an element of a set. (For example, it is an element of $Y$.) But the null set has no elements, and since $X=\emptyset$, $X$ has no elements and you cannot write $v \in X$ for any $v$ whatsoever, even $\emptyset$.

  2. $\emptyset \in Y$ because it was written that it is, as clearly as can be. The notation $Y=\{v\}$ means that $Y$ has one element, and $v$ is that element. Well, let $v=\emptyset$, that is, $Y=\{\emptyset\}$. The statement we made before about $v$ is now true about $\emptyset$: $Y$ has one element, and $\emptyset$ is that element.

  3. $Z$ has just one element. That element is the set $\{\emptyset\}$. But $\emptyset\neq\{\emptyset\}$. Therefore $\emptyset$ is not an element of $Z$.

  4. It is true that $X \subseteq Y$, but this is not because $\emptyset$ is an element of $Y$. It is because $X=\emptyset$, and $\emptyset$ is a subset of any other set that can ever be. In other words, it doesn't matter what is in $Y$.

  5. This is false. In fact, $Y \not\subseteq Z$, because $\emptyset \in Y$, but $\emptyset \not\in Z$. Again, $Z$ has just one element and that element is the set $\{\emptyset\}$, which is not the same thing as $\emptyset.$

  6. You are right, this is the same as 2. Since $X=\emptyset$, when we write $\emptyset\in Y$ we are saying that $X\in Y$.

  7. I don't know the mathematical definition of "directly within" as applied to sets. I suppose you mean that $\{\emptyset\}$ is found in the list of elements between the outer braces in the definition of $Z$, $Z=\{\ldots\}.$ So yes, $\{\emptyset\} \in Z$, and since $Y=\{\emptyset\}$, that implies $Y\in Z$.

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