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I learn set theory and took notes from a proof in a book, so that it makes sense to me. Can someone look if my notes/thinking is (formally and mathematically but most importantly conceptually) correct? Where does it look fishy or underexplained?

Theorem: Be $A$ a (dedekind-) infinite set. Then $A\setminus x$ is an infinite set too.

Proof: Be $A‘ \subset A$ and $f: A \to A‘$ bijective (since A is dedekind-infinite).

Because of $A‘ \subset A$ there is some $a \in A\setminus A‘$.

We assume $g: f$ with $dom(f) = A\setminus {a}$. $g$ is injective because $f$ is bijective and therefore also injective which isn‘t changed by the take away from $a$ from $f$‘s domain (but bijectivity might).

Now, $f(a) \notin rng(g)$ because $f(a) \in rng(f)$ and its preimage is $a \in A$, but such an element is impossible in $rng(g)$ because $g$ forbids a preimage $a \in A$.

Furthermore, $f(a) \neq a$ because of $rng(f) = A‘$.

Furthermore $a \notin rng(g)$ because $rng(g) \subseteq A‘$.

Furthermore and trivially $a \notin A \setminus a$, but $f(a) \in A \setminus a$, because $f(a) \neq a$ and else $rng(f) = A‘ \subset A$.

Because $rng(g) \subseteq A‘ \subset A$ and because $rng(g)$ misses $a, f(a)$ while $A \setminus a$ only misses $a$, we can conclude $rng(g) \subset A \setminus a$.

So is $g: f$ with $dom(f) = A \setminus a$, which can also be written $g: A \setminus a \to A‘$ or $g: A \setminus a \to rng(g)$, bijective?

Yes, because $g$ is injective (see above) and it‘s also surjective because of the very meaning of rng(g) to have at least a pre-image in its domain.

So we have $rng(g) \subset A \setminus a$ and $g: A \setminus a \to rng(g)$ bijective which makes $A\setminus a$ dedekind-infinite by Dedekind‘s definition.

But since $|A\setminus a| = |A\setminus x|$ where $x \in A$, it follows that $A \setminus x$ is dedekind-infinite. q.e.d.

This theorem even holds for any finite amount of n elements taken away from $A$ because you can just apply the proof technique iteratively: $A\setminus a$ can be proved to be dedekind infinite and so also $(A\setminus a)\setminus a$ and so on for $((A\setminus a) \setminus b)... \setminus n$.

Corrollary: Is $B$ (dedekind-) finite then so is $B \cup a$ because if $B \cup a$ was dedekind-infinite then our above theorem would make $B\setminus a$ (= $B$) infinite as well contrary to the assumption.

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  • $\begingroup$ Right at the start you have an error where you say $A' \subset A$ therefore there's a bijection between $A'$ and $A$. Consider $A =$ the integers, and $A = $ some finite set, say $\{3\}$. Secondly you assume $A'$ is a proper subset, but you didn't say that. Just a couple of minor issues but they occurred right away so I mentioned them. $\endgroup$ – user4894 May 30 at 5:59
  • $\begingroup$ An earlier issue is that neither your theorem statement nor the surrounding text ever indicate what "$a$" is. Without that information, the theorem is meaningless. $\endgroup$ – Paul Sinclair May 30 at 13:32
  • $\begingroup$ Guessing that $a$ is supposed to be an arbitrary element of $A$, another problem is that you limited $a \in A\setminus A'$, which is not true of all elements of $A$. $\endgroup$ – Paul Sinclair May 30 at 13:42
  • $\begingroup$ @Paul: I think I got the mistake and now my proof holds for any $x \in A$, showing $A \setminus x$ is infinite if A was infinite. $\endgroup$ – Pippen Jun 1 at 21:28
  • $\begingroup$ Understanding that English is not your native language, I try to be patient about a lot of things, but other than one word, "$g: f$ with $dom(f) = (A \setminus a \to A‘) = (A \setminus a \to rng(g))$" is mathematics, not English, but it is nonsensical mathematics. Please learn to treat "$=$" as a relation between mathematical objects, not as a general purpose connector. Stop abusing notations. Do not make your readers have to guess just what you could possibly mean by this mash. $\endgroup$ – Paul Sinclair Jun 2 at 0:15

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