Verify sup, max, min, inf for $\{\frac{n+3}{n}\;:\;n \in \mathbb N\}$

Question

I think I'm finally starting to understand this and wanted to check that I answered the following correctly:

$\{\frac{n+3}{n}\;:\;n \in \mathbb N\}$

inf: 1

min: 4

sup: DNE

max: DNE

Answer 1

Note that $n\mapsto(n+3)/n$ is decreasing. Moreover, $\lim_{n\rightarrow\infty}(n+3)/n=1$. Therefore, $\inf=1$.

For the same reasons, there is no $\min$. Indeed, suppose the minimum is attained at $n$. However, letting $m=n+1$, $(m+3)/m<(n+3)/n$, contradicting that $n$ is the point at which the minimum is attained.

Can you figure out the $\sup$/$\max$ case?

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Addendum: These definitions might be of help. In the following, $S\subset\mathbb{R}$ is arbitrary and nonempty.

The minimum of $S$ is an element $x\in S$ such that for all $s\in S$, $x\leq s$.

An infimum is defined differently. First, let's introduce the following:

A lower bound of $S$ is a number $x\in\mathbb{R}$ (this is the important part, $x$ is not necessarily restricted to be in $S$!) such that for all $s\in S$, $x\leq s$.

Now we can define infimum:

An infimum of $S$ is a lower bound $x$ of $S$ such that for all other lower bounds $x$ of $S$, $y\geq x$.

Maximums and supremums are defined similarly.