Does $\{π+1/n :\, n \in \Bbb N\}$ have an infimum?

Question

Does the set $$A=\left\{π+\frac1n :\, n \in \Bbb N\right\}$$ have infimum?

I assume $\pi$ is the infimum, because when $n$ tends to infinity the smallest possible element of $A$ tends to $\pi$, does not hit the $\pi$. I do not know if I am right, or if I am right, how should prove what the infimum of this set is?

Thanks.

Answer 1

You are right. Since $(\forall n\in\mathbb N):\pi\leqslant\pi+\frac1n$, $\pi$ is a lower bound of $A$. And, since $\lim_{n\to\infty}\pi+\frac1n=\pi$, $\pi=\inf A$.

Answer 2

Your set contains the number $\pi+\frac1{43}$, so it is not empty.

All of its elements are positive, so it is bounded below by $0$.

A nonempty set of reals that is bounded below always has an infimum. Case closed.

Answer 3

Completing the thoughts from Henning and José.

$\pi < \pi+\frac1n$ for any natural $n$, so $\pi$ is a lower bound.

Next, let $\varepsilon > 0$, and let $m = 1 + \left\lceil \frac{1}{\varepsilon} \right\rceil$. $m$ is a natural number, so $\pi + \frac1m$ is in the set, but $\pi + \frac1m < \pi + \varepsilon$, so $\pi + \varepsilon$ is not a lower bound of the set.

Therefore $\pi$ is the infimum of the set.