I just asked a question to know if a proof using "or" to make a justification was valid logically.
Here is an example of my confusion.
Show that if $A$ is an infinite subset of $X$, then $\bar A=X$ in the finite complement topology.
Here was my proof. Let $x \in X, U$ be an open set containing $x$.If $U=X$, then $U \cap A \neq \varnothing$. Suppose $U \neq X$. Then $U=X-S$ for some finite set $S$. Consider $A \cap (X-S)$. Then this intersection is nonempty (here is where I am unsure if this is correct) otherwise $A$ or $X-S$ would have to be finite.(A would be finite since $A \subset S$ if the intersection is empty, however I noticed the second statement is not necessarily true). So $U \cap A \neq \varnothing$ for each open set $U$, and since $x$ was arbitrary, $\bar A=X$.
So the statement "otherwise,…" is my justification for the contradiction that will arise if $(X-S) \cap A = \varnothing$. So I want to know if including the not necessarily true statement $X-S$ would have to be finite, ruins the validity of my proof.
Best Answer
Yes, the assertion that
is technically true, because in fact one can conclude that $A$ must be finite, but in ordinary mathematical writing it will be understood to mean that there are two ways in which $A\cap(X\setminus S)=\varnothing$ can be true, $A$ finite or $X\setminus S$ finite, that both can occur, and that they exhaust the possibilities. But the fact that $A$ is infinite does not imply that its complement is finite, so the fact that $A$ and $X\setminus S$ are disjoint doesn’t mean that at least one must be finite.
Fortunately, the argument can be fixed quite easily. Let $x\in X$, and let $U$ be an open nbhd of $x$. Then $U=X\setminus S$ for some finite $S\subseteq X$, so
$$A\cap U=A\cap(X\setminus S)=(A\cap X)\setminus S=A\setminus S\ne\varnothing\,,$$
since $A$ is infinite.
Note that there is no need to consider the case $U=X$ separately: that’s simply $U=X\setminus S$ for $S=\varnothing$.