Subset of a set of sets

The elements of set A are

$a,b,\{\{c,d\},e\}$

and the subsets of set A are

$\{\},\{a\},\{b\},\{\{\{c,d\},e\}\},\{a,b\},\{a,\{\{c,d\},e\}\},\{b,\{\{c,d\},e\}\},\{a,b,\{\{c,d\},e\}\}$ $


No, it is not a subset. sets are packages of things and when it comes the the existential idententy a set is a different thing than the elements within its packaging.

$A = \{a, b, \{\{c,d\},e \}\}$ has $3$ elements. The elements listed are:

  • $a$
  • $b$
  • $\{\{c,d\},e\}$.

And the set $\{c,d\}$ has $2$ elements. They are:

  • $c$
  • $d$

For $\{c,d\}$ to be a subset of $A$ every one of the two elements of $\{c,d\}$ must be one of the three elements of $A$. That is not the case.

$c$ is not equal to $a$ or to $b$. And $c$ does not equal $\{\{c,d\},e\}$.

Yes, $c$ is an element of $\{c,d\}$. And $\{c,d\}$ is an element of $\{\{c,d\},e\}$. And finally $\{\{c,d\},e\}$ is an element of $A$, but that's three levels of removal. And being an element of a set does not transfer, not even a single level.

....

As $c$ is not an element of $A$ then not every element of $\{c,d\}$ is an element of $A$ so $\{c,d\}$ is not a subset $A$. End of story.

There in no need to check whether $d$ is an element of $A$ as... End of story means end of story.... but if we were to check if $d$ is element of $A$ we'd find it is not for the exact same reason.

=====

Your question does lead to valid question: Is there are term for the concept of a "descension" of membership of a set. If $a$ is an object, and $A$ is a set we say $a$ is an element of $A$ if .... $a$ is an element of $A$. Is there are term for if there is a set $A_1$ that is an element of (not a subset of) and $a_1$?

For example if $A= \{a,b,\{c,d\}, \{\{e,f,\{g,\{h,i\}\}\},w\}\}$ the $a\in A$ but $c \not \in A$ but $A_1 = \{c,d\}$ and $A_1\in A$ and $c\in A_1$. Is there term for this?

If we called it $c \in_2 A$ we could define that if $x \in_{k+1} A$ if there is a set $A_k$ so that $A_k\in_k A$ and $x \in A_k$.

So for example $e\in \{e,f,\{g,\{h,i\}\}\}$ and $\{e,f,\{g,\{h,i\}\}\} \in \{\{e,f,\{g,\{h,i\}\}\},w\}$ and $ \{\{e,f,\{g,\{h,i\}\}\},w\}\in A$ so $e \in_2 A$.

And $g\in_3 A$. And $h\in_4 A$.

Is there a term for this concept?

.... Maybe but if so I don't know it. But in terms of being an element of a set. They don't transfer. They just don't


To emphasize with the use of color, your set $A$ is the set:

$$A=\{\color{green}{a},\color{blue}{b},\color{red}{\{\{c,d\},e\}}\}$$

The elements of $A$ are written between the outermost brackets and are separated by commas. Each individual element above is distinctly colored so as to emphasize that the entirety of the expression for each element is necessary in describing the element as a whole.

In particular, notice how the red element in the above, although itself being a set containing another set and more elements within it... the element is considered a single object.

Now... if we want to ask about whether $\{\color{purple}{c},\color{orange}{d}\}$ is a subset of $\{\color{green}{a},\color{blue}{b},\color{red}{\{\{c,d\},e\}}\}$ that would require the purple element on the left $\color{purple}{c}$ to be equal to (not just a part of) one of the elements of $A$ as well as the same for the orange element $\color{orange}{d}$.

However, the purple element $\color{purple}{c}$ is not equal to any of the three elements, $\color{green}{a},\color{blue}{b},\color{red}{\{\{c,d\},e\}}$ and so $\{c,d\}$ fails to be a subset of $A$.

Tags:

Elementary Set Theory

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