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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?

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  • $\begingroup$ 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. $\endgroup$ Sep 24 '20 at 21:03
  • $\begingroup$ 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. $\endgroup$
    – Doug M
    Sep 24 '20 at 21:04
  • $\begingroup$ Similar question $\endgroup$ Sep 24 '20 at 21:05
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    $\begingroup$ "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. $\endgroup$
    – user239203
    Sep 24 '20 at 21:05
  • $\begingroup$ @Gae.S.And it should be noted that the answer may be yes! $\endgroup$ Sep 24 '20 at 21:13
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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 $3 \in X$?

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    $\begingroup$ 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. $\endgroup$
    – fleablood
    Sep 24 '20 at 21:21
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    $\begingroup$ 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 $3$. $\endgroup$
    – User203940
    Sep 24 '20 at 21:23
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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 $\{3\}\subset \{2,3,4\}$.

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

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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

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

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  • $\begingroup$ 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.) $\endgroup$
    – fleablood
    Sep 24 '20 at 21:59

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