Formally put: If $M$ is an infinite set, and $E$ a finite subset $E\subset M$, then $|M\setminus E| = |M|$ (cardinality is unchanged when removing a finite number of elements).
It is readily obvious that
$$
(M\setminus E) \cup E = M
$$
Since $(M\setminus E)$ and $E$ are obviously disjoint, then the following rule applies:
$$|(M\setminus E) \cup E| = |(M\setminus E)| + |E|$$
Inserting this on the left hand side of the above…
$$|(M\setminus E)| + |E| = |M|$$
Since $E$ was finite, it must have some finite cardinality number $k\in\mathbb{N}$, such that
$$|(M\setminus E)| + |E| = |(M\setminus E)| + k = |M| \iff |(M\setminus E)| = |M| – k$$
And since $|M|$ was an infinite set, subtracting any natural number from its cardinal number does not change it. Thus the $-k$ may be omitted, and the proof is done.
This was just the best I could cobble together from some basic set theory and reasoning. But I've seen some uses on the internet – without references! – of a rule like $|A\cup B| = max\{|A|,|B|\}$ which applies in the case of infinite sets? I haven't been able to find a reference for this, but I can see how it would be used here.
Any way, is the above sufficient? Is the rule I used assured to work here?
Best Answer
What you need to do, by definition, is find a bijection from $M$ to $M\setminus E.$ The idea that you take away a finite number of elements from an infinite set and it can't stop it from being infinite is not a bad heuristic, but how do you know it doesn't decrease its cardinality to a smaller, but still infinite, cardinality? You are effectively assuming what needs to be proven.
The way we establish a bijection is we use the fact that such a bijection can be readily obtained for a countably infinite set. So let $M_0$ be a countably infinite set such that $E\subseteq M_0\subseteq M$ and then prove there is a bijection between $M_0$ and $M_0\setminus E,$ which can then be extended to a bijection between $M$ and $M\setminus E.$
The existence of such an $M_0$ actually requires the axiom of choice, so this whole affair is surprisingly nontrivial. To tie this to what I said initially, it is really quite literally the case that in the absence of the axiom of choice it is consistent that $M\setminus E$ could have a strictly smaller cardinality from $M$ (though still infinite).
More detail on the two steps outlined above.
Note, the second argument uses dependent choice, but this is overkill: the statement every infinite set has a countably infinite subset is actually strictly weaker than countable choice, and it's not too hard to find a proof using only countable choice.