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SOLUTION

Given the question in the question tab, the following are the solution steps to solve the question

STEP 1: Define a countably infinite set

A set is countably infinite if its elements can be put in one-to-one correspondence with the set of natural numbers. In other words, one can count off all elements in the set in such a way that, even though the counting will take forever, you will get to any particular element in a finite amount of time.

STEP 2: Write the given set

[tex]\mleft\lbrace2q\colon q\in Q+\mright\rbrace[/tex]

This set is read as a set of values 2q such that q is an element of positive rational numbers

STEP 3: Show the proof

[tex]\begin{gathered} \text{Let }Q_{\pm}=\mleft\lbrace q\in Q\colon q>0\mright\rbrace \\ \text{For ev}ery\text{ q}\in Q_+,\text{ there exists at least one pair (m,n)}\in N\times N \\ \text{such that q}=\frac{m}{n} \end{gathered}[/tex]

Therefore, we can find an injection:

[tex]i\colon Q_+\Rightarrow N\times N[/tex]

By Cartesian Product of Natural Numbers with Itself is Countable, N×N is countable.

Hence Q+ is countable, by Domain of Injection to Countable Set is Countable.

Note that the function defined by is a bijection. Therefore the composition is a bijection. So the given set is countably infinite.

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