# Problem

Given sets $$\mathcal A$$, $$\mathcal B$$, $$\mathcal Y$$, let $$\mathcal X$$ be a set with the following properties:

• $$\mathcal X$$ $$\supset$$ $$\mathcal A$$ and $$\mathcal X$$ $$\supset$$ $$\mathcal B$$,
• if $$\mathcal Y$$ $$\supset$$ $$\mathcal A$$ and $$\mathcal Y$$ $$\supset$$ $$\mathcal B$$ , then $$\mathcal Y$$ $$\supset$$ $$\mathcal X$$

Prove that $$\mathcal X$$ = $$\mathcal A$$ $$\cup$$ $$\mathcal B$$.

# My work

Let $$\mathcal x$$ be an element of set $$\mathcal X$$ and $$\mathcal y$$ be an element of set $$\mathcal Y$$.

If $$\mathcal X \supset\mathcal A$$, then $$\mathcal x$$ $$\in$$ $$\mathcal A$$, then $$\mathcal x$$ $$\in$$ $$\mathcal X$$. If $$\mathcal X \supset\mathcal B$$, then $$\mathcal y$$ $$\in$$ $$\mathcal B$$, then $$\mathcal y$$ $$\in$$ $$\mathcal Y$$.

If $$\forall$$ $$\ x$$ $$\in$$ $$\mathcal A$$, $$\ x$$ $$\in$$ $$\mathcal X$$; and $$\forall$$ $$\ y$$ $$\in$$ $$\mathcal B$$, $$\ y$$ $$\in$$ $$\mathcal Y$$, then $$\mathcal X = \mathcal A \cup \mathcal B$$.

This is an introductory problem in a Real Analysis course I'm doing on my own. I'm an Engineer trying to acquire a deeper mathematical maturity. The book I'm following doesn't bring answers, ergo, the question. But more than a "Right or Wrong" answer, I'd like an evaluation concerning the rigor - or the lack thereof - of the answer I provided.

• Just a light hearted comment: who uses caligraphic symbols for sets? Isn't it a pain to type these? Aug 5, 2019 at 23:35
• hahah. It is indeed! Aug 5, 2019 at 23:39

The thing is that $$A,B, X$$ are specific sets that exist and a given to you. They do not change.

$$Y$$ can be any set in the world.

So let $$Y = A\cup B$$.

(We know in general that $$A \subset A\cup B$$ and $$B\subset A\cup B$$. This isn't just true of these sets. This would be true for any sets, $$K$$ and $$M$$. It is always true that $$K \subset K\cup M$$. That is because $$K\cup M$$ is the set of all elements that are in $$K$$ or in $$M$$. If $$x \in K$$ then it is one of the elements that are in $$K$$ or in $$M$$. So all the elements of $$K$$ are elements in $$K\cup M$$. So $$K\subset K\cup M$$.)

$$A\subset A\cup B$$ or in other words $$Y= A\cup B \supset A$$. And $$B\subset A\cup B$$ or in other words $$Y= A\cup B \supset B$$.

But we have a condition that if $$Y\sup A$$ and $$Y\sup B$$ then we will have $$Y\sup X$$.

And we have do have that $$A\cup B \supset A$$ and $$A\cup B \supset B$$ so we must have $$A\cup B\supset X$$.

On the other hand: $$A\cup B \subset X$$. We know this because if $$x \in A\cup B$$ then either $$x \in A$$ or $$x\in B$$. If $$x \in A$$ then, because $$X\sup A$$ so every element in $$A$$ is in $$X$$, we know $$x \in X$$. And if $$x \in B$$ then $$x\in X$$ because $$X\sup B$$ and that's what $$X\sup B$$ means. So either way $$x \in X$$.

So every element in $$A\cup B$$ is in $$X$$ so $$A\cup B \subset X$$.

We also have $$A\cup X \supset X$$ so every element in $$X$$ is in $$A\cup B$$.

So $$A\cup B$$ and $$X$$ both have the exact same elements.

So $$A\cup B = X$$.

....

"x be an element of set X... If X⊃A, then x ∈ A,then x ∈ X"

"If $$X \subset A$$" No if about it! You were told that $$X \supset A$$ so nothing to speculate.

"then $$x \in A$$". That's not true. Just pick an $$x$$ that is in $$X$$ but not in $$A$$. Then $$x$$ will not be in $$A$$.

"then x ∈ X" But... you already said that.

... and so on.

Your proof is not correct. There is no $$\mathcal Y$$ to begin with so your first statement doesn't make sense.

First verify that $$\mathcal A \cup \mathcal B \subset \mathcal X$$. This is easy from the first condition and the definition of union.

Next let us prove that $$\mathcal X \subset \mathcal A \cup \mathcal B$$. Let us prove this by contradiction. Suppose $$x \in \mathcal X \setminus \mathcal A \cup \mathcal B$$. Take $$\mathcal Y=\mathcal A \cup \mathcal B \setminus \{x\}$$. Use the second condition given to show that we must have $$\mathcal X \subset\mathcal A \cup \mathcal B \setminus \{x\}$$. Arrive at a contardiction by observing that $$x$$ belongs to LHS but not to RHS.

• Thank you for your answer. I noticed, thanks to your comment, that I had forgotten the Y in the first line of my problem statement. Then, I edited the question. Aug 5, 2019 at 23:42
• How do you read the slash in $\mathcal X$ \ $\mathcal A \cup \mathcal B$? Aug 5, 2019 at 23:43
• I used slash to to denote set difference: $A \setminus B$ is the set of all points which belong to $A$ but not to $B$. Aug 5, 2019 at 23:47
• " I noticed, thanks to your comment, that I had forgotten the Y in the first line of my problem statement. Then, I edited the question." What you edited no longer makes any sense. It only made sense without specifying the $Y$. Consider $X = \{a,b,c,d,e\}$ and $A=\{a\}$ and $B=\{b\}$ and $Y = \{a,b,c,d,e,f,g$. Then $X\supset A$ $X\supset B$, $Y\supset A$, $Y\supset B$ and $Y\supset X$. So all conditions are true. But $A \cup B = \{a,b\} \ne \{a,b,c,d,e\} = X$. Aug 6, 2019 at 0:08