Contradiction of axioms of real numbers

Question

I am just starting out in real analysis, so please bare with me. My questions concerns three specific properties of the real numbers, at least as far as i understand them. Those are:

1. The natural numbers $\mathbb{N}$ are a subset of the real numbers $\mathbb{R}$.
2. The real numbers $\mathbb{R}$ have the completeness axiom.
3. The archimedean property holds.

Here is where I struggle:

By the completeness axiom, any subset of $\mathbb{R}$ has a supremum. Since $\mathbb{N}$ is a subset of $\mathbb{R}$, $\mathbb{N}$ should have a supremum. But by the archimedean property, that is not true. From this i come to the following (probably wrong) conclusions:

1. The natural numbers $\mathbb{N}$ are not a subset of the real numbers $\mathbb{R}$ or
2. The real numbers $\mathbb{R}$ do not have the completeness axiom.

How is this "contradiction" resolved? Do i have a wrong idea of what being a subset means or what the completeness axiom means?

Any help is appreciated.

Cheers
Maxwell

Edit: As was pointed out, i've got the completeness axiom wrong. The natural numbers $\mathbb{N}$ do not satisfy it.

Answer 1

First of all, it is good that you are actively stress-testing the assumptions you've been told! That's exactly what you should be doing, and that instinct will serve you well throughout mathematics.

In this case, however, the issue is that you've misinterpreted one of the assumptions. Completeness says the following:

Every nonempty set of reals which is bounded above has a least upper bound.

Both the bolded points are crucial:

• $\emptyset$ has lots of upper bounds, but no least upper bound since there is no least real number.

• No unbounded-above set of real numbers - such as $\mathbb{N}$ (as you observe) or $\mathbb{R}$ itself (more simply) - can have a least upper bound, obviously.

And it's the second point that's relevant to your question.