# Is $\{3\}$ a subset of $\{\{1\},\{1,2\},\{1,2,3\}\}$?

Is $$\{3\}$$ a subset of $$\{\{1\},\{1,2\},\{1,2,3\}\}$$?

If the set contained $$\{3\}$$ plain and simply I would know but does the element $$\{1,2,3\}$$ include $$\{3\}$$ such that it would be a subset?

• No, it's not. Let $A=\{\{1\},\{1,2\},\{1,2,3\}\}$. The subsets must themselves contain sets, since A is a set of sets. Then $\{\{1\}\}$ is a subset of A, $\{\{1\},\{1,2\}\}$ is another subset of A. $\{\{1,2\},\{1,2,3\}\}$ is also a subset of A. $\{3\}$ is a subset of an element of A. Sep 24 '20 at 21:03
• No, when you have a set of sets, you do not "look through" the brackets. The sets are members, but the members of those sets are not members. Sep 24 '20 at 21:04
• Similar question Sep 24 '20 at 21:05
• "Is $3=\{1\}$?" ; "Is $3=\{1,2\}$?" ; "Is $3=\{1,2,3\}$?": If the answer to one of these questions is yes, then $\{3\}$ is a subset of $\{\{1\},\{1,2\},\{1,2,3\}\}$; otherwise, it isn't.
– user239203
Sep 24 '20 at 21:05
• @Gae.S.And it should be noted that the answer may be yes! Sep 24 '20 at 21:13

Let $$X$$ be a set. We say $$Y \subseteq X$$ ($$Y$$ is a subset of $$X$$) if, for all $$x \in Y$$, we have $$x \in X$$.

Examine the sets $$Y = \{3\}$$, $$X = \{\{1\},\{1,2\},\{1,2,3\}\}$$. Take $$x = 3 \in Y$$. Is $$3 \in X$$?

Trickier problem: If $$X = \{\{1\},\{3\},\{1,2\},\{1,2,3\}\}$$, is $$3 \in X$$?

• Why is the trickier problem trickier? $3\not \in X$ because $3\ne \{1\};3\ne \{3\},3\ne \{1,2\}, 3\ne \{1,2,3\}$. Why is that trickier. Sep 24 '20 at 21:21
• I would expect someone to say "Oh, I see 3 by itself" and say it is in the set. I wanted to emphasize the difference between $\{3\}$ and $3$. Sep 24 '20 at 21:23

No.

the elements inside elements of the set do not count.

The elments of your big set are:

• $$\{1\}$$
• $$\{1,2\}$$
• $$\{1,2,3\}$$.

The elements of your small set are:

• $$3$$

So $$\{3\}$$ is a subset only if $$3$$ is equal (the same thing; !!!!!NOT!!!! an element within) one of the elements $$\{1\}$$ or $$\{1,2\}$$, or $$\{1,2,3\}$$. But none of those are the same thing as $$3$$ so $$\{3\}$$ is not a subset.

But in some text the natural numbers are defined as

$$0 = \emptyset$$

$$1= \{\emptyset\}$$

$$2= \{\emptyset, 1\}$$.

$$3 = \{\emptyset, 1, 2\}$$

So we could have a trick thing of $$\{3\} \subset \{\{\emptyset,1\}, \{\emptyset,1,2\}, \{\emptyset,1,2,3\}\}$$ not because $$3 \in \{\emptyset,1,2,3\}$$ (that's utterly irrelevent), but because $$3 = \{\emptyset, 1, 2\}$$ and the set $$\{\{\emptyset,1\}, \{\emptyset,1,2\}, \{\emptyset,1,2,3\}\}$$ is equally equal to the set $$\{2,3,4\}$$ and $$\{3\}\subset \{2,3,4\}$$.

No it is not, we have that $$\{1\}\in \{\{1\},\{1,2\},\{1,2,3\}\}$$ but $$\{3\}$$ is not an element of the set, what is true is that $$3\in\{1,2,3\}$$.

As you can see, there are plenty of good answers here.

I think that the most important thing to understand is:

$$3 \neq \{ 3 \}.$$

These are two different objects.

Moreover the followings hold:

$$3 \in \{3\}$$ $$3 \in \{\ldots, 3, \ldots \}$$ $$\{3\} \in \{\ldots, \{3\}, \ldots\}$$ and

$$\{3\} \not\in \{3\}$$ $$\{3\} \not\in \{\ldots, 3, \ldots \}$$

• In for a penny..... might as well add (and perhaps germaine to the common misconception) $3\not \in \{........., \{3\},......\}$ (unless it is one of the objects omitted in the "..."s.) Sep 24 '20 at 21:59