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$x$ represents an element, not a set.

Assume $S:=\{X \cup x\}$ is countably finite. This means that there is a bijection between $S$ and a subset of $\mathbb{N}$. This is clearly not the case since $X$ is infinite. I don't really know how to write down that last part rigorously. How can I approach this?

Thanks in advance.

MyWorld
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    You mean $x \notin X$, not $X \notin x$. Also, presumably $X \cup {x}$ rather than ${X \cup x}$. – Robert Israel Jan 20 '19 at 19:39
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    Very important remark. Even if $X$ is really huge, ${X}$ is very very small. So small that it only has one element, $X$. Hence your set $S$ is countably finite (or simply finite) and it is in bijection with the subset of natural numbers consisting in just one element. The bijection is $X\cup x\longmapsto n\in\mathbb N$. – Dog_69 Jan 20 '19 at 19:39
  • I did not think about that! Thank you for your thoughtful remark. So if I was to consider $X \cup x$, I could say it is countably infinite, right? How could that be proved? – MyWorld Jan 20 '19 at 19:44
  • @RobertIsrael I edited the question, thanks for noticing. However, I do mean ${X \cup x}$. – MyWorld Jan 20 '19 at 19:46
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    @Zachary If $x$ is countable, yes. Maybe you meant $X\cup{ x}$? – J.G. Jan 20 '19 at 19:47
  • or perhaps you mean $x \not \subset X$ or perhaps $X \cap x = \emptyset$ – Henry Jan 20 '19 at 20:02

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As the comments point out, your notation is a little confusing. The set $S = \lbrace X\cup x\rbrace$ contains one element: $X\cup x$. In particular, $S$ is countable. On the other hand, the set $X\cup x$ may not be countable, even if $X$ is countable. This is because $x$ could be uncountable and disjoint from $X$. The statement you are probably looking for is the following:

Proposition. If $X$ is countable and $x\not\in X$, then $X\cup \lbrace x\rbrace$ is countable.

In fact, something much stronger is true. If $X_1,X_2,\dots $ is a sequence of countable sets (indexed over the natural numbers), then $\bigcup_i^\infty X_i$ is countable. To prove this, use that $\mathbb N\times\mathbb N$ is countable and construct a surjection $\mathbb N\times\mathbb N\rightarrow \bigcup_i^\infty X_i$. The proposition above is then the special case where $X_1 = X$, $X_2 = \lbrace x\rbrace$ and $X_i = \varnothing$ for $i > 2$.

Addendum. Now how would you prove that $\mathbb N\times\mathbb N$ is countable? Here it may help to visualize $\mathbb N\times \mathbb N$ as the nodes of an infinite grid in the first quadrant of $\mathbb R^2$. A natural way to "count" these nodes is to "spiral outwards".

o.h.
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  • How to prove that $f:\mathbb N\times\mathbb N\to \bigcup_i^\infty X_i$ is bijective? – PNT Oct 31 '21 at 13:01