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