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For all $n \in \mathbb{N}$, we have $n\geq1$, which implies that 1 is the infimum of $\mathbb{N}$. Indeed, $E=\{n \in \mathbb N : n\geq1$} contains 1 by definition of $\mathbb{N}$. In addition, if $n \in \mathbb{E}$, then $n +1 \in \mathbb{N}$ because $n \in \mathbb{N}$, with n+1 > n+0=n $\geq1$. So, $n +1 \in \mathbb{E}$. We conclude that the infimum of the naturals is 1 because 1 is a minorant of the naturals which belongs to the naturals.

The only part which I'm not sure of understanding is how did they establish the inequalities. with n+1 > n+0=n $\geq1$

We know that 1>0, and we can add n to both sides to have n+1>n but how do you get the n superior or equal to 1 ?

Thanks

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    $n\geq 1$ by definition of the natural numbers. – 2017-01-15
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    So, just to be clear, when we "create" the inequality n+1>n, we can establish that n is superior or equal to 1 because the definition says so, right ? – 2017-01-15
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    Right.${}{}{}{}$ – 2017-01-15
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    Thanks, can you put an answer so I can accept you ? – 2017-01-15

1 Answers 1

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$n\geq 1$ by definition of the natural numbers.

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    not completely correct: by definition $n+1\geq n$ and $(a\geq b \wedge b\geq c)\rightarrow a\geq c$, easily implying $n\geq 1$ by induction. – 2017-01-15
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    The symbols used in my book is strictly superior to n... n+1>n+0=n>=1 – 2017-01-15