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  • Show that If $X$ is an any set, then $X\subset X\cup\left\{ X\right\}$

Proof. Let $t\in X$. We must show $t\in X\cup\left\{ X\right\}$, that is we need to show either $t\in X$ or $t\in\left\{ X\right\}$, so we know that $t\in X$, hence we are done.

Can you check my proof?

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    $\begingroup$ Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but". $\endgroup$ – DonAntonio Dec 25 '18 at 0:31
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    $\begingroup$ Ehm. Well. You're perfectly right. But the question unsettles me a little. :D $\endgroup$ – YoungMath Dec 25 '18 at 0:31
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    $\begingroup$ @YoungMath the questin was from my exam. I think It is very trivially but I see that the question don't says that $X\subseteq X\cup\left\{X\right\}$, from this do we get any problem? $\endgroup$ – pozcukushimatostreet Dec 25 '18 at 0:35
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    $\begingroup$ @KathySong Yes, $\subset$ makes a difference. You need to show $\exists a\in X\cup\{X\}$ such that $a\notin X$. That $a$ is $X$ $\endgroup$ – Shubham Johri Dec 25 '18 at 0:41
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    $\begingroup$ Oh well, that's subtile, you're right. Then, you need to show that $X \neq X \cup \{X\}$. However, this is simple since $X \notin X$ due to the axiom of regularity. $\endgroup$ – YoungMath Dec 25 '18 at 0:42
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Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X \subseteq X \cup \{X\}$ is already discussed in full detail, I want to make a remark about the case $X \neq X \cup \{X \}$.

The Axiom of Regularity reads $$ \forall x \left( x \neq \emptyset \Rightarrow \exists y \in x: y \cap x = \emptyset \right).$$

Let us prove, that $X \notin X$ for all sets $X$.

Soo, let $X$ be any set. Due to the axiom of pairing, $\{X\}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X \cap \{X\} = \emptyset$ since $X$ is the only element in $\{X\}$. Hence, $X \notin X$.

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Exercise.
Show for all sets A and B, that A $\subseteq$ A $\cup$ B.

With that, your problem is just a special case.

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  • $\begingroup$ Thanks for exercise. I would like to ask a question that how can I show $X\neq X\cup\left\{X\right\}$ Can you help? $\endgroup$ – pozcukushimatostreet Dec 25 '18 at 1:50
  • $\begingroup$ @KathySong. Equality would lead to X in X in contradiction to axiom of regularity. $\endgroup$ – William Elliot Dec 25 '18 at 1:58
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    $\begingroup$ Please read the comments before posting an answer. Thank you! :) $\endgroup$ – YoungMath Dec 25 '18 at 2:57

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