proving the infimum of a set

Question

I am trying to find the supremum and infimum of the set $A = \{1/x +(-1)^x|x \in \mathbb{N} \}$.

Obviously, the supremum is $3/2$, however, I am not sure about the infimum. In the limit as $x \rightarrow \infty$, $1/n \rightarrow 0$ and since $(-1)^x$ is always either $-1$ or $1$, I would think the infimum should be $-1$, but I may be confusing the definition of a lower bound and an infimum.

The infimum is the greatest lower bound, so $-1$ is indeed correct — Henry, Oct 28 '20 at 10:50

score 1 · Accepted Answer · answered Oct 28 '20 at 10:48

You are correct, the infimum is indeed $-1$. $-1$ is also a lower bound of $A$.

You can prove that $-1=\inf(A)$ by proving two things:

That $-1$ is a lower bound for $A$ (this one should be easy)
That $-1$ is the largest lower bound for $A$, in other words, that for all $\epsilon > 0$, the value $-1+\epsilon$ is not a lower bound for $A$.

proving the infimum of a set

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