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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 $A + 2$?

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

More generally, we have:

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).

Try to prove this theorem as an exercise.

For your reference: Suppose $X$ is infinite and $A$ is a finite subset of $X$. Then $X$ and $X \setminus A$ are equinumerous

and Suppose $X$ is infinite and $A$ is a finite subset of $X$. Then $X$ and $X\setminus A$ are equinumerous

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