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I desperately need clarifications on the least upper bound property (as the title suggests). Here's the main question:

Why doesn't the set of rational numbers ℚ satisfy the least upper bound property?

Every textbook/website answer I have found uses this example:

Let S={x∈ℚ : x≤√2}. This has an upper bound in ℚ, but it has no supremum in ℚ, since √2 is irrational, therefore ℚ does not satisfy the least upper bound property.

Here's my major problem:

A supremum is not necessarily an element of the set it's being a least upper bound of.

By the definition of the least upper bound property:

(i)A real number x is called an upper bound for S if x ≥ s for all s ∈ S.

(ii)A real number x is the least upper bound (or supremum) for S if x is an upper bound for S and x ≤ y for every upper

bound of S.

Nowhere does this definition state that the supremum has to belong in that set.

It seems to me that √2 should perfectly qualify as the supremum of S by this very definition. After all x≤√2 for all x∈S.

But why do we dismiss it by saying it is not rational? Why does it HAVE to be? Why does the supremum of a subset of ℚ also HAS to be in ℚ, or else ℚ doesn't satisfy the least upper bound property?

I'm probably missing something and this question looks idiotic to some, but frankly I don't see it. I hope someone can help me out by giving a detailed, comprehensible answer. Thanks in advance.

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# Clarifications on the least upper bound property and the irrational numbers

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