Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality of $T$ is $\aleph$? Let $T =\{X\in P(\mathbb{Q}) | X\cup \mathbb{N} = \mathbb{Q} \}$ the cardinality  of $T$ is $\aleph$ ? prove
attempt
we know that $X\in P(\mathbb{Q})$ so, $X\cup \mathbb{N} = \mathbb{Q}$ create an infinity numbers
 A: It is clear from your definition that $T = \{ A \cup (\mathbb Q \setminus \mathbb N) \, | \, A \in P(\mathbb N) \}$. It can obviously be put in bijection with $P(\mathbb N)$ via the bijection $f: T \to P(\mathbb N)$ that sends $X$ to $X \cap \mathbb N$.
Hope that helps,
A: It helps to get a "picture" of what is going on.
$T = \{X \in P(\mathbb Q)| ......\}$ means that $T$ is a set of subsets of $\mathbb Q$.  Our question is just which of the subsets and how many of subsets are in $T$.
If $X$ is one of the elements of $T$, the $X$ is subset of $T$ with the following property:  $X\cup \mathbb N = \mathbb Q$.
Now as $\mathbb N$ contains no non-integer rationals, doesn't contain $0$, and no negative rationals, $X$ must "provide" those in order for $X\cup \mathbb N = \mathbb Q$.  And if $L = \{q\in \mathbb Q| q\not \in \mathbb N\}=\mathbb Q\setminus L$ then $L\subset X$ for all $X\in T$.
So what else is in $X$?  That's the choice we have.  Every $X\in T$ is a subset of $\mathbb Q$ that contains $L$ as a subset.  And $X$ may or may not contain some natural numbers.  Lets define $X_a = \{n\in \mathbb N| n= X\}=$ the set of all natural numbers in $X$.
Then $X = L \cup X_a= $(the set of all rational numbers that are not in $\mathbb N$) $\cup$ (the set of all natural numbers in $X$).
Now we can describe each of the $X \in T$ by identifying which naturals it contains (everything else it has will be had by everty other $Y\in T$ so we can take those for granted).
So if we define $f(X) =X_a$ that is a mapping for $X \in T$ to $X_a$ which is in $P(\mathbb N)$ as it is subset of $\mathbb N$.  So $f: T\to P(\mathbb N)$.
And further more $f$ is a bijection.  It's injective because for any two different $X,Y\in T$ they both will contain all the non-naturals so the only thing that distinguishes them are the natural numbers they contain.  In other words, $X \ne Y$ if and only if $X_a \ne Y_a$.
And $f$ is surjective.  For any $K \subset N$ then the set $L \cup K$ will one of the subsets in $T$.
So $f:T\to P(\mathbb N)$ is a bijection.
So $|T| = |P(\mathbb N)|$.
.....
Now I think in you class you have probably had it proven that $|P(\mathbb N)| = 2^{\mathbb N} = |\mathbb R| \ne  |\mathbb N|$.
so $T =  |\mathbb R| \ne  |\mathbb N|$.
