The subset of the positive reals has no smallest element

Question

There is statement in the book:

"The subset $ A =\{\frac{1}{n} : n \in \mathbb{N} \} $ of the positive reals has no smallest $ n $ element because for any $ x_0 = \frac{1}{n} \in A $ that we might pick, there is always a smaller element $ \frac{1}{n+1} \in A $."

And the book said it is false.

But I found similar questions (first, second, third) that prove the statement.

Is the book wrong and the statement is truth?

Answer 1

Adding a few more words to the paragraph in question, added text in bold.

The well-ordering principle looks simple, but it actually makes a profound and fundamental statement about the positive integers $\Bbb N$, that it and every subset of it has a smallest element. In fact, the corresponding statement about the positive real numbers, that it and every subset of it has a smallest element, is false: For illustrative purposes we can prove that there is no smallest element for the subset $A=\{1/n~:~n\in\Bbb N\}$ of the positive reals because for any $x_0=\frac{1}{n}\in A$ that we choose, there's always a smaller element $1/(n+1)\in A$. Similarly, we can prove that no smallest positive real number exists by noting that for any positive real number $\epsilon$, we can choose a smaller positive real number $\frac{\epsilon}{2}$.

The text was not wrong, but was perhaps unclear to you what "the corresponding" statement was in reference to. The corresponding statement was what preceded the sentence in question, that there was a smallest element. You incorrectly interpreted "the corresponding statement" to be in reference to the statement that followed which was a different statement.

The text, talking about the set $A=\{1/n~:~n\in\Bbb N\}$ also technically proves nothing about the real numbers. The proof that the positive real numbers has no minimum I included at the end of the paragraph.

Answer 2

Yes. The book is wrong. The sequence $$\{a_n\}=\left\{1,{1\over 2},{1\over 3},{1\over 4},{1\over 5},\cdots\right\}$$is strictly decreasing and has no minimum. Same is true for $A$.

Answer 3

Actually, the book says that that statement is true. What it says that it is false is that the set $\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$, endowed with the usual order $\leqslant$, is well-ordered.