Proving $-1$ is the infimum of $E=\{\frac{2}{n}+(-1)^n\mid n\in \mathbb{N}\}$

Question

Prove: $\inf\{E\}=-1$ for $E=\{\frac{2}{n}+(-1)^n\mid n\in \mathbb{N}\}$

Let assume the contrary , there is $x\in E$ such that $\frac{2}{n}+(-1)^n\leq-1$.

Because we are looking at negative numbers we can look only at odd $n\in \mathbb{N}$ that is $\frac{2}{n}-1\leq-1\Rightarrow \frac{2}{n}\leq0$

But for all $n\in \mathbb{N}$ $\frac{2}{n}>0$, therefore $\inf\{E\}=-1$

Is the proof vaild?

Answer 1

No. As it stands, what's written in the proof is correct (although worded in a somewhat confusing manner), but it doesn't actually prove the claim. All you prove is that $-1$ is a lower bound of $E$, but that doesn't imply that $-1$ is actually the infimum. After all: What you've written works perfectly well if you replace $-1$ with $-2$. You need to also show that there is no lower bound of $E$ larger than $-1$.

---

To rewrite it, here are some suggestions:

• State clearly that you're proving that $-1$ is a lower bound. This doesn't need to be done by contradiction - just do it directly.

• Now you need to prove that $-1$ is the greatest lower bound. To this end, take a number $x > -1$. Find an odd value of $n$ so that $$-1 < -1 + \frac 2 n < x$$ which can be done by algebra. Then explain how this shows the claim.

Answer 2

"Let assume the contrary , there is $x\in E$ such that $\frac{2}{n}+(-1)^n\leq-1$."

That is incorrect. The fact that some member of $E$ is $\le-1$ does not in any way contradict the statement that $-1$ is the infimum of $E$. What you need to show is that every member of $E$ is $\ge-1$ and there is no number $A>-1$ such that every member of $E$ is $\ge A$.

Answer 3

Hmm so I tried my hand at this since it'd be good review.

To show that -1 is inf E, we need to show that:

$\forall u\in U$(where u is the group of all the lower bounds of E) $\forall e\in E$ this occurs:

$$u\lt -1 \lt e$$

Let's prove by contradiction. Assume $inf(E) \ne -1$. Then there must exist a n that the following holds true for odd numbers and even numbers:

$$\frac{2}{n}+(-1)^n \lt -1$$

Let's try for odd numbers:

$$\frac{2}{n}-1\lt-1 \rightarrow \frac{2}{n}\lt0$$

This is a contradiction, no n will make this inequality true.

Let's try for even numbers:

$$\frac{2}{n}+1\lt-1 \rightarrow \frac{2}{n}\lt-2$$

This won't hold true either. $n\gt0$ and $2\gt0$. Therefore we proved that $\forall e \in E $

$$-1\lt e$$

Now we need to show that $u \le n$. Since we assumed $infE\ne-1$, then we can use the lower bound property which states that if a group is lower bounded and non-empty that it has an infimum, which we can define as inf(e)=a. Now I think you can show Think you can take over from here?