Is $\pi$ in the infinite set $ \{ 3, 3.1, 3.14, 3.141, 3.1415, ...\} $?

Question

Does $\pi$ exist in the following infinite set. Apologies for lack of set notation, and i'm hoping its not necessary to help me understand the nature of infinite decimal expansion.

I know sets are unordered but for simplicity and intuition I will show the set being generated in increasing size.

$$ \{ 3, 3.1, 3.14, 3.141, 3.1415, ...\} $$

The definition of the set is 'An infinite set containing $\pi$'s decimal representation from 0 to $\infty$ significant figures.'

My (very limited!) intuition based on the notion of infinity and decimal expansion is that this set cannot contain $\pi$.

However, intuition based on how infinite sequences rather than sets can converge hints the opposite. Is it possible that a sequence generated this way will converge towards $\pi$ and an infinite set generated this way will never contain $\pi$?

If this question is daft and can be answered with a simple axiom relating to irrational numbers, please be kind, i'm more hobbyist than math student. Thanks.

Answer 1

All the numbers in your sequence are decimal numbers, which are a subset of the rational numbers. But $\pi$ is not a rational number, therefore it is not a member of this sequence.

Answer 2

First off, great question!

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To answer your question, no, $\pi$ will not be present in that infinite set, because all numbers in it are numbers with terminating decimals (in other words, they are rational numbers). However, $\pi$ is not rational, so it cannot be in the set.

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To say the same thing in another manner, assume for the sake of contradiction that $\pi$ exists in that set. We do know that $\pi$ has infinite digits and that it is not rational. But all elements of that infinite set are rational (because they have terminating decimals), and hence we have a contradiction.

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I hope this helped! If you have any questions regarding my explanation, please feel free to ask in the comments :)