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Is $\mathbb{N}\cup \big\{\sqrt{2}\big\}$ an uncountable set?

I think it is.

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    $\begingroup$ Before answering, do you think $\mathbb N$ is countable? $\endgroup$ – peter a g Nov 9 '17 at 18:09
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    $\begingroup$ Great. Write the first few elements of $\mathbb N$ on one line, and write directly below those items, $\sqrt 2$ first, and then the next few elements of $\mathbb N$. Hope this helps! $\endgroup$ – peter a g Nov 9 '17 at 18:11
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    $\begingroup$ @Stu "what is strange, one element make it uncountable" No, quite the opposite. Thinking about Hilbert's hotel may help give you a sense of how the cardinality of an infinite set behaves when you add things to it ... $\endgroup$ – Noah Schweber Nov 9 '17 at 18:15
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    $\begingroup$ It's so strange that it is untrue. $\endgroup$ – Lee Mosher Nov 9 '17 at 18:16
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    $\begingroup$ My "Great" comment above was unclear, and even misleading. Below, in the meantime, others have been clearer, but pride requires that I save face: "... $\sqrt 2$ first, and then the FIRST few elements of $\mathbb N$." $\endgroup$ – peter a g Nov 9 '17 at 18:25
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No, it is not. Since we can create bijection between $\mathbb{N} \cup \{ \sqrt{2}\}$ and $\mathbb{N}$.

Let $f: \mathbb{N} \cup \{ \sqrt{2}\} \to \mathbb{N}$ be given by:

$f(1) = \sqrt{2}$ and $f(n) = n-1$ for $n \ge 2$.

It is clear that $f$ is a bijection, which means the two sets must be equal in cardinality. Therefore both sets are countable.

Note that the set that results from adding any finite number of points (or even a countable number of points) to a countable set is still countable.

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  • $\begingroup$ Last question [0,1] is R is uncountable, isn't it? $\endgroup$ – Stu Nov 9 '17 at 18:16
  • $\begingroup$ Yes, $[0,1]$ is uncountable. $\endgroup$ – Joel Nov 9 '17 at 18:17
  • $\begingroup$ The reals are uncountable. They do have important countable subsets. Indeed, the rational numbers are a countable subset of $\mathbb{R}$, and the rational numbers are dense in $\mathbb{R}$. Simply adding one irrational number to $\mathbb{Q}$ doesn't make the new set uncountable. The point here is that there are so many more irrational numbers than rational numbers that we cannot get the collection of real numbers by simply adding one at a time. $\endgroup$ – Joel Nov 9 '17 at 18:19
  • $\begingroup$ You might also be interested to know that the set of all algebraic numbers (which includes $\sqrt{2}$, $\sqrt[3]{7}$, $\phi = \frac{1+\sqrt{5}}{2}$ etc.) is also countable. $\endgroup$ – Joel Nov 9 '17 at 18:20
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No, it isn't. You can start counting from $\big\{\sqrt{2}\big\}$ and then proceed to counting $\mathbb{N}$

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It is countable. Here is an explicit enumeration of its elements:

$a_1 = \sqrt{2}$

$a_2 = 1$

$a_3 = 2$

$a_4 = 3$

$a_5 = 4$

$\cdots$

$a_n = n-1$ (for $n>1$)

$\cdots$

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If you add countably many elements to a countable set then the resulting set is also countable. Here $\Bbb N$ is countable and we are adding a single element $\sqrt 2$ to it. Thus the resulting set is countable.

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Consider : $A:=$ {$\mathbb{N}$}$\bigcup ${$√2$} ;

$A$ as the union of 2 countable sets is countable.

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