Is it true if the set X is countably infinite then there are bijective maps f: X-->N and f:N-->X ?
Separately, could someone start me off on this proof:
Prove that if A in uncountable, then A - {x} is uncountable.
Thanks
Is it true if the set X is countably infinite then there are bijective maps f: X-->N and f:N-->X ?
Separately, could someone start me off on this proof:
Prove that if A in uncountable, then A - {x} is uncountable.
Thanks
Suppose that $\displaystyle A$ was uncountable but $\displaystyle A-\{x\}$ was countable. Then, there exists a bijection $\displaystyle f:A-\{x\}\mapsto\mathbb{N}$ and so $\displaystyle g:A\mapsto \mathbb{N}\cup \{0\}$ given by $\displaystyle g(z)=\begin{cases} f(z) & \mbox{if} \quad z\ne x \\ 0 & \mbox{if}\quad z=x\end{cases}$. This is clearly a bijeciton and thus a contradiction.
In fact, there is a much stronger statement that can be made.
Let $\displaystyle E$ be uncountable and $\displaystyle F\subseteq E$ be countable. Then $\displaystyle E-F$ is uncountable.
Proof: Since $\displaystyle E-F$ is countable there exists some $\displaystyle f:E-F\mapsto \mathbb{N}$ which is bijective. Also, since $\displaystyle F$ is countable there is some $\displaystyle g:F\mapsto \mathbb{Z}-\mathbb{N}$ which is also bijective. Clearly then, $\displaystyle \eta:E\mapsto\mathbb{Z}$ given by $\displaystyle \eta(x)=\begin{cases} f(x) & \mbox{if} \quad x\in E-F \\ g(x) &\mbox{if} \quad x\in F\end{cases}$ is a bijection, this of course contradicts that $\displaystyle E$ is uncountable.