Why can it consist of only one element of a?

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The discussion centers on the concept of singleton sets in set theory, specifically addressing why the set notation $\{a\}$ represents a set containing only one element, $a$. It emphasizes that $\{a\}$ is equivalent to $\{a,a\}$ and illustrates this through the axiom of extensionality, which states that two sets are equal if they contain the same elements. The conversation also touches on the distinction between sets and multisets, highlighting that $\{1, 1\}$ is not equal to $\{1\}$ in certain structures.

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evinda
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Hey! (Nerd)

If $a,b$ sets, then we symbolize $\{ a,b\}$, the set that has as elements $a$ and $b$ and only these.

If $\{ a, b \}$ a set of pair and $a=b$, then we write $\{a,a \}=\{a \}$ and the set $\{ a \}$ is called the singleton, that consists of only one element of $a$.

Why can $\{a\}$ consist of only one element of $a$ ? (Thinking)
 
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evinda said:
Why can $\{a\}$ consist of only one element of $a$ ?
Strictly speaking, it is not the case that $\{a\}$ consists of only one element of $a$; rather, $\{a\}$ consists of only one element, namely, $a$.

By definition of $\{a,b\}$, we have
\[
x\in \{a,b\}\leftrightarrow x=a\lor x=b.\qquad(*)
\]
Also by definition, $\{a\}=\{a,a\}$. Instantiating (*) with $\{a\}$ (i.e., when $a=b$), we get
\[
x\in \{a\}\leftrightarrow x=a\lor x=a\leftrightarrow x=a.
\]
 
Evgeny.Makarov said:
Strictly speaking, it is not the case that $\{a\}$ consists of only one element of $a$; rather, $\{a\}$ consists of only one element, namely, $a$.

By definition of $\{a,b\}$, we have
\[
x\in \{a,b\}\leftrightarrow x=a\lor x=b.\qquad(*)
\]
Also by definition, $\{a\}=\{a,a\}$. Instantiating (*) with $\{a\}$ (i.e., when $a=b$), we get
\[
x\in \{a\}\leftrightarrow x=a\lor x=a\leftrightarrow x=a.
\]

I understand... Thank you very much! (Clapping)
 
Note that:

$P \vee P \iff P$

"If I am going to the store, or, I'm going to the store, then, I'm going to the store. Moreover, if I'm going to the store, then surely, either I'm going to the store, or...I'm going to the store!"

Note as well, that:

$\{a,a\} = \{a\} \cup \{a\}$

More pointedly: "sets are not multisets", or as G. Spenser-Brown says, in The Laws of Form (paraphrased):

To call again, is to call.
 
This is actually known as the axiom of extensionality. It is defined that two sets $A$ and $B$ are equal if every element of $A$ is in $B$. From this follows that $\{1, 1\}$ for example is equal to the singleton set $\{1\}$.

However, this is only an axiom. There are set-like structures for which this axiom doesn't hold, e.g., $\{1, 1\}$ and $\{1\}$ differs. A multiset is an example of such a structure, for example.

EDIT : Ah, Deveno beat me to it.
 
mathbalarka said:
This is actually known as the axiom of extensionality. It is defined that two sets $A$ and $B$ are equal if every element of $A$ is in $B$. From this follows that $\{1, 1\}$ for example is equal to the singleton set $\{1\}$.

However, this is only an axiom. There are set-like structures for which this axiom doesn't hold, e.g., $\{1, 1\}$ and $\{1\}$ differs. A multiset is an example of such a structure, for example.

EDIT : Ah, Deveno beat me to it.

Do we get extra credit for thinking of the same thing, at the same time?
 
Deveno said:
Do we get extra credit for thinking of the same thing, at the same time?

Hahaha, probably not. We'll be marked as copy-cats by the examiners soon enough.
 

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