# Show that $X\subset X\cup\left\{ X\right\}$

• 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?

• Is so simple that any more explanation just would make it clumsy. It is fine. A little thing: change that "so" by "but". – DonAntonio Dec 25 '18 at 0:31
• Ehm. Well. You're perfectly right. But the question unsettles me a little. :D – YoungMath Dec 25 '18 at 0:31
• @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? – pozcukushimatostreet Dec 25 '18 at 0:35
• @KathySong Yes, $\subset$ makes a difference. You need to show $\exists a\in X\cup\{X\}$ such that $a\notin X$. That $a$ is $X$ – Shubham Johri Dec 25 '18 at 0:41
• 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. – YoungMath Dec 25 '18 at 0:42

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, $$X \notin X$$.

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

With that, your problem is just a special case.

• 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? – pozcukushimatostreet Dec 25 '18 at 1:50
• @KathySong. Equality would lead to X in X in contradiction to axiom of regularity. – William Elliot Dec 25 '18 at 1:58