Altering an Infinite Set does not change cardinality Let X be an infinite set. Show that adding or subtracting a single point does not change its cardinality. 
I have a plan but need help writing the actual proof. 
I need to show that it doesn't matter which point is removed, and then I can use the fact that X is in one-to-one correspondence with a proper subset to prove this. 
 A: Hint: show that there exists an injective map $\omega\to X$. Then it is possible to "hide" a single point by shifting $\omega$ up/down.
A: HINT: Given two elements $x,y\in X$ there is a permutation of $X$, $\pi$, such that $\pi(x)=y$ and $\pi(y)=x$.
A: Let $X$ be the infinite set, $Y \subset X$ the set for which there's a bijection $Y \rightarrow X$ (which means $|Y|=|X|$), and $x$ some element in $X$.
Since there's at least one element "missing" in $Y$: 
$$|X|=|Y| \leq |X- \{x\}|$$
Using the same reasoning:
$$|X- \{x\}| \leq |X|$$
Conclusion:
$$|X- \{x\}| = |X|$$  
A: Actually, it can be proven in ZFC that adding or removing an element from an infinite set never changes its cardinality but it cannot be proven in ZF. In ZFC, it can be proven as follows. From the axiom of choice, we can derive the axiom of dependent choice. Using the axiom of dependent choice, we can show that every infinite set has a countably infinite subset. For any infinite set, since it has a countably infinite subset, adding or removing an element doesn't change its cardinality.
What really is cardinality? ZFC doesn't really formally define the cardinality of any set. However, we can informally define the cardinality of any set to be the smallest ordinal number such that there's a bijection from that set to that ordinal number. That's what the mathematical community decided they mean when they talk about the cardinality of a set in ZFC.
