\(\inf\big(\{\frac{1}{n}: n \in \mathbf{N}\} \big) = 0.\)


For the definitions of an upper bound and the least upper bound of a set, see this.
For the Archimedean Principle, see this

Proof:

Let \(A = \{\frac{1}{n}: n \in \mathbf{N}\}\). To prove that \(\inf(A)=0\), we will show that 0 is a lower bound of \(A\) and that for all \(\epsilon > 0\), \(0 + \epsilon\) is not a lower bound of \(A\). First, since \(n \in \mathbf{N}\) is positive and 1 is positive, then \(1/n > 0\) for all \(n \in \mathbf{n}\) and so \(0\) is a lower bound on \(A\).

Next, to show that \(0\) is the greatest lower bound, we need to find an element \(x \in A\) such that for any \(\epsilon > 0\), \(x\) will be lower than \(0 + \epsilon\). To find such element we know by the Archimedean principle that there exists an \(n \in \mathbf{N}\) such that \(\epsilon \geq \frac{1}{n}\). So just choose \(x = \frac{1}{n}\) since we are working in \(\mathbf{N}\). Re-arranging

$$ \begin{align*} \frac{1}{n} &\leq \epsilon \\ x &\leq 0 + \epsilon. \end{align*} $$

From this we see that 0 is the greatest lower bound on \(A\). \(\blacksquare\)

References: