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I'm a math teacher and I have a doubt about this question

$\{x\}\in\{x,\{x,y\}\}$ is true or false? And why?

I think it's true because $\{x,y\}=\{x\}\cup\{y\}$ then...

But I'm not sure. Thanks!

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Concisely, $\{ x\}$ is a set, $x$ is an element. –  Nameless Mar 18 at 23:32

5 Answers 5

up vote 2 down vote accepted
  1. "Belongs to" ($\in$) means simply "is an element of".
  2. The set $\{x,\{x,y\}\}$ has just two elements, $x$ and $\{x, y\}$, and neither one is $\{x\}$. (To phrase it differently: neither element is equal to $\{x\}$.)*
  3. So $\{x\}$ does not belong to $\{x,\{x,y\}\}$.

$^*$Footnote: Certainly not for arbitrary $x$ and under standard set theory, as the question intends.

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If $x=y$ then $\{x\}=\{x,x\}=\{x,y\}\in\{x,\{x,y\}\}$.

If $x=\{x\}$ then also yes, because $\{x\}=x\in\{x,\{x,y\}\}$.

Otherwise, the answer is no. There is no element in $\{x,\{x,y\}\}$ which is exactly $\{x\}$. It is true that $\{x\}$ is a subset of $\{x,\{x,y\}\}$ and a subset of an element of this set as well. But it is not an element itself.

The general method I often suggest is to use some dummy variable. Set $u=\{x,y\}$ and $v=\{x\}$. Then $u\neq v$ because $y\in u\setminus v$. Then we ask, $v\in\{x,u\}?$ and the answer is quite obvious, $v\neq x$ and $v\neq y$. So $v\notin\{x,u\}$.

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You need to understand the difference between $x$ and $\{x\}$. For example $\emptyset$ is just an empty set but $\{\emptyset\}$ is a set containing one element $\emptyset$. Hence the answer to your question is NO as RHS contains $x$ but not $\{x\}$.

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Good example! Just a minor friendly suggestion: Use \emptyset for $\emptyset$, \phi ($\phi$) is quite different. (Or perhaps I'm too much of a stickler.) –  Utku Alhan Mar 18 at 18:51
you're right, thanks :) –  wanderer Mar 18 at 18:53

It is not true since $\{x\}$ is not an element of $\{x, \{x,y\}\}$.

What is true is that $x \in \{x, \{x,y \}\}$. It is also true that $x\in \{x, \{x,y\}\}$ and that $\{x,y\}\in \{x, \{x,y\}\}$ because $\{x, \{x,y\}\}$ consists of exactly the two elements $x$ and $\{x,y\}$.

So in general you want to only say that something is in a set if exactly that something is on the list of things between the $\{$ and $\}$.

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By the definition of set enumeration notation, $$\{ x \} \in \{ x, \{ x, y \} \}$$ if and only if $$ \{ x \} = x \qquad \text{or} \qquad \{ x \} = \{ x, y \} $$

For the second case, $$\{ x \} = \{ x, y \} $$ if and only if $$ \{ x \} \subseteq \{ x, y \} \qquad \text{and} \qquad \{ x \} \supseteq \{ x, y \} $$ if and only if (since the first case is always true we only need to check the second case) $$ x \in \{ x \} \qquad \text{and} \qquad y \in \{ x \} $$ if and only if (since the first case is always true we only need to check the second case) $$ y = x$$

Therefore, $$\{ x \} \in \{ x, \{ x, y \} \}$$ if and only if $$ x = \{ x \} \qquad \text{or} \qquad y = x $$ ...

Depending on your particular version of set theory (e.g. "axiom of foundation"), $x = \{ x \}$ can never be true. In that case, we can follow up with

... if and only if $y = x$

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