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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.

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    $\begingroup$ Just a light hearted comment: who uses caligraphic symbols for sets? Isn't it a pain to type these? $\endgroup$ Aug 5, 2019 at 23:35
  • $\begingroup$ hahah. It is indeed! $\endgroup$
    – Jxson99
    Aug 5, 2019 at 23:39

2 Answers 2

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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 $A\cup B = X$.

....

As to your work:

"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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jxson99
    Aug 5, 2019 at 23:42
  • $\begingroup$ How do you read the slash in $\mathcal X$ \ $\mathcal A \cup \mathcal B$? $\endgroup$
    – Jxson99
    Aug 5, 2019 at 23:43
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    $\begingroup$ I used slash to to denote set difference: $A \setminus B$ is the set of all points which belong to $A$ but not to $B$. $\endgroup$ Aug 5, 2019 at 23:47
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    $\begingroup$ " 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 $A \cup B = \{a,b\} \ne \{a,b,c,d,e\} = X$. $\endgroup$
    – fleablood
    Aug 6, 2019 at 0:08

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