# Dedekind-infinite set stays infinite even if losing one element

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.

• 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. – user4894 May 30 at 5:59
• 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. – Paul Sinclair May 30 at 13:32
• 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 $A$. – Paul Sinclair May 30 at 13:42
• @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. – Pippen Jun 1 at 21:28
• 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. – Paul Sinclair Jun 2 at 0:15