# Cardinality and subsets

Let $$A$$ be a proper infinite subset of some set $$X$$. If $$x, y$$ are two distinct elements of $$X$$ that are not in $$A$$, we may set $$B = \{x, y\} \cup A$$. What is the cardinality of $$B$$ in terms of the cardinality of $$A$$? Justify your answer.

It's probably wrong but if $$B$$ is the union of $$\{x,y\}$$ and $$A$$, then isn't the cardinality of $$B$$ just the cardinality of $$A + 2$$?

• If $A$ is infinite, so is $B$. – Sean Roberson Oct 2 '18 at 2:29
• $A$ is an infinite subset. Adding a finite number of elements to an infinite set leaves the cardinality unchanged. Do you see why? – Dave Oct 2 '18 at 2:30
• Yes I do, I glazed over the fact that A is an infinite subset. Thanks! – Amanda Varvak Oct 2 '18 at 2:31
• real-analysis?? – Alex Kruckman Oct 2 '18 at 2:33
• Consider $A=\mathbb{Q}\subset\mathbb{R}=X$ and take $\{\sqrt{2},\sqrt{3}\}$ as $\{x,y\}$. – Sujit Bhattacharyya Oct 2 '18 at 2:34

If $$X$$ is infinite and $$Y$$ is a finite subset of $$X$$, then $$X$$ and $$X\setminus Y$$ are equinumerous (or equivalently, have the same cardinality).
The gist of the above theorem lies in the fact that If $$X$$ is infinite, then there exists $$B\subseteq X$$ such that B is countably infinite (Here we assume Axiom of Countable Choice).