Prove that $\mathbb{Q}$ and $\mathbb{Q} \cup \{\pi, e\}$ have the same cardinality.
I know I must show that there exists a bijection between these two sets but I'm having a difficult time trying to come up with a function that relates them. Any suggestions? Thanks.
Hint: Let $f: \mathbb{Q} \to \mathbb{Q} \cup \{\pi, e\}$ be the function $$ f(x) = \begin{cases} \pi&\text{if } x = 0 \\ e &\text{if } x = 1 \\ x-2 &\text{if } x \in \{2, 3, 4, 5, 6, \ldots\} \\ x &\text{otherwise}. \end{cases} $$
Here's an outline of a proof:
Let $f$ be defined as
$f(n)=n+2$ for $n\in\mathbb{N}$ and $f(x)=x$ for $x\in \mathbb{Q}-\mathbb{N}$.
Prove that $f$ is bijective from $\mathbb{Q}$ onto $\mathbb{Q}-\{1,2\}$. Let now $g:\mathbb{Q}\cup \{\pi,e\}\to \mathbb{Q}$ be defined as: $g(x)=f(x)$ if $x\in \mathbb{Q}$, and $g(e)=0, g(\pi)=1$. Prove that $g$ is bijective and then you're done.
Well, in theory, if you have shown $\mathbb Q $ is countable, we have a bijection $j:\mathbb N\rightarrow \mathbb Q $.
We create a new bijection, k, where $k (1)=\pi; k (2)=e;\forall n>2| k (n)=j (n-2) $.
Of course if we want to add two (or any finite number of elements) elements to an uncountable set it should be easy but seems hard.
Other folks have shown this with the reals. Generalized it's...
What we do (we can assume countable choice, can't we?) Is arbitrarily pick a countable subset $P=${$x_1,x_2... $}. Then we can define bijective $j|X \rightarrow X \cup\{a,b\}$ as $j (x_1)=a;j (x_2)=b$ for $x_i\in P; i >2;j (x_i)=x_{i-2}$ and for $x \not \in P;j (x)=x $.
Union of Infinitely countable and finitely countable set, is again Infinitely countable. Meaning same cardinality. You could add any number of finite Irrational numbers, not just pi and e.
