Find the Infimum and Supremum of the Set

Let

$$S=\left\{(-1)^n+\frac{1}{m}: n\in\mathbb{N},\ m\in\{1,2,3,\ldots\}\right\}\cup\{-1\}.$$

Thus $m$ ranges over the positive integers, while $n$ ranges over the natural numbers. The key is to split the expression according to whether $n$ is even or odd. That determines the sign of $(-1)^n$ and makes the bounds visible.

For this set, both the infimum and supremum are actually elements of the set: the infimum is $-1$, and the supremum is $2$. Here is how to verify the bounds and check that each is attained.

Separate the even and odd cases

If $n$ is even, then $(-1)^n=1$, so the expression becomes

$$1+\frac{1}{m}.$$

As $m$ ranges over the positive integers, these values are $2, 3/2, 4/3, 5/4,\ldots$. Each is greater than $1$, and the largest is $2$, obtained at $m=1$.

If $n$ is odd, then $(-1)^n=-1$, and the expression becomes

$$-1+\frac{1}{m}.$$

For positive $m$, these values are $0,-1/2,-2/3,-3/4,\ldots$. Each is greater than $-1$, and the values approach $-1$ as $m$ grows. The set also includes $-1$ explicitly, through the separate term $\{-1\}$.

Thus the set consists of the values from these two cases together with the explicitly included point $-1$.

Verify the infimum

For every positive integer $m$, $1/m>0$. Therefore, in the odd case,

$$-1+\frac{1}{m}>-1.$$

In the even case, $1+1/m>1$, so those values are also greater than $-1$. The additional element $-1$ is a lower bound for the whole set and belongs to the set itself. Hence it is the least element, and therefore also the infimum:

$$\inf S=-1.$$

The fact that the odd-case values approach $-1$ is consistent with this result, but it is not needed to establish the infimum once we know that $-1$ is included and every other value is at least $-1$.

Verify the supremum

For even $n$, the expression is $1+1/m$. Since $m\geq 1$, we have $1/m\leq 1$, and so

$$1+\frac{1}{m}\leq 2.$$

For odd $n$, the expression $-1+1/m$ is at most $0$, so it is also less than $2$. The extra element $-1$ is less than $2$ as well. Thus $2$ is an upper bound for the set.

To show that it is the least upper bound, check whether the set reaches it. Taking $m=1$ and any even $n$ gives

$$(-1)^n+\frac{1}{m}=1+1=2.$$

So $2$ belongs to the set. It is the maximum, and therefore the supremum:

$$\sup S=2.$$

A few values as a check

Taking $n=1$ (odd) and $n=2$ (even), and using $m=1,2,3$, gives:

Parity of $n$	$m=1$	$m=2$	$m=3$
Odd	$0$	$-1/2$	$-2/3$
Even	$2$	$3/2$	$4/3$

These values show both the descending odd-case values and the even-case values that start at $2$. The separate $-1$ in the definition lies below all the listed values.

Common mistakes

• Ignoring the parity of $n$. The term $(-1)^n$ is not always $1$ or always $-1$; both cases occur as $n$ varies.
• Forgetting the explicitly included $-1$. The odd-case values approach $-1$ from above, but do not equal it for any positive $m$. The separate element makes $-1$ part of the set.
• Assuming the supremum is not attained. Although some sequences of values approach a bound without reaching it, here $m=1$ in the even case gives $2$ exactly.
• Confusing an infimum or supremum with a minimum or maximum. In this example both bounds are attained: $-1$ is the minimum and $2$ is the maximum. In other sets, a bound need not belong to the set.

If a text defines $\mathbb{N}$ to include $0$, the expression $1/m$ still requires $m\ne 0$. Interpreting $m$ as a positive integer gives the same result under either convention for whether $0$ belongs to $\mathbb{N}$, because both even and odd values of $n$ occur.