# Supremum and Infimum of 1/n+1/m

#### Alfredo Dawlabany

##### New member
We have the set $$\displaystyle E=\left \{ \frac{1}{n}+\frac{1}{m},m,n\in \mathbb{N^*} \right \}$$
My teacher solved it in this way:

$$\displaystyle \forall n,m,\frac{1}{n}+\frac{1}{m}\leqslant 1+1\leqslant 2$$
Then 2 is an upper bound of E and since 2 belongs to E for n=1 and m=1 then 2=sup E (sup E is done)

$$\displaystyle \forall n,m\in \mathbb{N^*},\frac{1}{n}+\frac{1}{m}> 0\Rightarrow 0$$ is a lower bound of E
but $$\displaystyle 0\notin E$$
(for now everything is clear)
But after that we have to see if 0 is the greatest lower bound so that 0=inf E .
Here my teacher said that we have to prove that $$\displaystyle \forall \varepsilon >0$$ , $$\displaystyle \exists p\in E$$ such that $$\displaystyle 0<p<0+\varepsilon$$
and that $$\displaystyle \exists n_0,m_0\in \mathbb{N^*}$$ such that $$\displaystyle 0<\frac{1}{n_0}+\frac{1}{m_0}<\varepsilon$$
Then the proof went like this
$$\displaystyle \forall \varepsilon >0,\exists n_0$$ such that $$\displaystyle \frac{1}{n_0}<\frac{\varepsilon }{2}$$
$$\displaystyle \forall \varepsilon >0,\exists m_0$$ such that $$\displaystyle \frac{1}{m_0}<\frac{\varepsilon }{2}$$
$$\displaystyle \Rightarrow \frac{1}{n_0}+\frac{1}{m_0}< \varepsilon \Rightarrow 0+\varepsilon$$ is not a lower bound of E (contradiction)
then 0=inf E
Now, the problem is that I didn't know why we have to prove this and also the I didn't understand the proof

#### ksdhart2

##### Senior Member
I'll address your two main concerns in order. First, "Why do we need to show that 0 is the greatest lower bound?"

At the point where the proof stops making sense to you, we've shown that every element of the set E is greater than 0. Thus, it is a lower bound. But can we be sure there isn't some other number, greater than 0, such that every element of E is also greater than that number? Suppose for a moment the sequence was $$\displaystyle F = \left\{\dfrac{1}{2n} \bigg \rvert n \in \mathbb{N}^* \right\}$$. As before, it's definitely true that every element of F is greater than 0, so 0 is a lower bound. However, n must be a strictly positive real number. If we plug in n = 1, we can see that the smallest possible element in F is 1/2. This means that 0 isn't the infimum of F, even though it is a lower bound.

Second, "What is the proof doing? I don't understand it."

To show that 0 is the greatest lower bound, and thus the infimum, we need to show that, if we pick any positive number (this is the $$\displaystyle \epsilon$$), there is an element of E that's bigger than 0 but less than our choice of $$\displaystyle \epsilon$$. What your professor did in the proof was to split up the sequence defining E into two separate sequences. We know that $$\displaystyle \dfrac{1}{n}$$ races off towards 0 as n increases towards infinity, and the same applies for $$\displaystyle \dfrac{1}{m}$$. Thus, by picking a big enough value for n, we can guarantee that $$\displaystyle \dfrac{1}{n} < \epsilon$$, no matter how small an $$\displaystyle \epsilon$$ we pick. In fact, since n can be as big as we need, $$\displaystyle \dfrac{1}{n}$$ can be made even smaller than $$\displaystyle \dfrac{\epsilon}{2}$$.

Say we picked $$\displaystyle \epsilon = 10^{-33} = 0.000000000000000000000000000000001$$. That's super duper crazy small! But, even so, we can still pick a big enough n to satisfy the criteria. Specifically, we can pick, say, $$\displaystyle n = 10^{100}$$, because $$\displaystyle \dfrac{1}{n} = \dfrac{1}{10^{100}} = 10^{-100} \ll \dfrac{10^{-33}}{2}$$

Then, because both $$\displaystyle \dfrac{1}{n}$$ and $$\displaystyle \dfrac{1}{m}$$ can be made smaller than $$\displaystyle \dfrac{\epsilon}{2}$$, we can definitely guarantee that their sum is smaller than $$\displaystyle \epsilon$$. And thus, for any choice $$\displaystyle \epsilon > 0$$, there exists an element of E smaller than it, and so we've proven there cannot exist a lower bound of E except 0. We therefore know that 0 is the infimum of E.

#### Alfredo Dawlabany

##### New member
I'll address your two main concerns in order. First, "Why do we need to show that 0 is the greatest lower bound?"

At the point where the proof stops making sense to you, we've shown that every element of the set E is greater than 0. Thus, it is a lower bound. But can we be sure there isn't some other number, greater than 0, such that every element of E is also greater than that number? Suppose for a moment the sequence was $$\displaystyle F = \left\{\dfrac{1}{2n} \bigg \rvert n \in \mathbb{N}^* \right\}$$. As before, it's definitely true that every element of F is greater than 0, so 0 is a lower bound. However, n must be a strictly positive real number. If we plug in n = 1, we can see that the smallest possible element in F is 1/2. This means that 0 isn't the infimum of F, even though it is a lower bound.

Second, "What is the proof doing? I don't understand it."

To show that 0 is the greatest lower bound, and thus the infimum, we need to show that, if we pick any positive number (this is the $$\displaystyle \epsilon$$), there is an element of E that's bigger than 0 but less than our choice of $$\displaystyle \epsilon$$. What your professor did in the proof was to split up the sequence defining E into two separate sequences. We know that $$\displaystyle \dfrac{1}{n}$$ races off towards 0 as n increases towards infinity, and the same applies for $$\displaystyle \dfrac{1}{m}$$. Thus, by picking a big enough value for n, we can guarantee that $$\displaystyle \dfrac{1}{n} < \epsilon$$, no matter how small an $$\displaystyle \epsilon$$ we pick. In fact, since n can be as big as we need, $$\displaystyle \dfrac{1}{n}$$ can be made even smaller than $$\displaystyle \dfrac{\epsilon}{2}$$.

Say we picked $$\displaystyle \epsilon = 10^{-33} = 0.000000000000000000000000000000001$$. That's super duper crazy small! But, even so, we can still pick a big enough n to satisfy the criteria. Specifically, we can pick, say, $$\displaystyle n = 10^{100}$$, because $$\displaystyle \dfrac{1}{n} = \dfrac{1}{10^{100}} = 10^{-100} \ll \dfrac{10^{-33}}{2}$$

Then, because both $$\displaystyle \dfrac{1}{n}$$ and $$\displaystyle \dfrac{1}{m}$$ can be made smaller than $$\displaystyle \dfrac{\epsilon}{2}$$, we can definitely guarantee that their sum is smaller than $$\displaystyle \epsilon$$. And thus, for any choice $$\displaystyle \epsilon > 0$$, there exists an element of E smaller than it, and so we've proven there cannot exist a lower bound of E except 0. We therefore know that 0 is the infimum of E.
Thank you I understood it now more
But I was thinking that if we can suppose that any number "p"
is a lower bound of E such that p>0 and then prove by contradiction that p <0 so that 0=inf E .
Can we do this ? If yes , how could we proceed ?

#### ksdhart2

##### Senior Member
Thank you I understood it now more
But I was thinking that if we can suppose that any number "p"
is a lower bound of E such that p>0 and then prove by contradiction that p <0 so that 0=inf E .
Can we do this ? If yes , how could we proceed ?
Absolutely. In fact, that's essentially what the proof does. Whether you call the number you pick p or $$\displaystyle \epsilon$$ is of no consequence. If we begin by assuming that there exists some p > 0 that is a lower bound of E, what we're really saying is that:

$$\displaystyle \forall n \in \mathbb{N}^* \: \forall m \in \mathbb{N}^* \: \left( \dfrac{1}{n}+\dfrac{1}{m} \ge p \right)$$

We can then reformulate this in terms of limits:

$$\displaystyle \displaystyle \lim_{(n,m) \to (\infty, \infty)}\left( \dfrac{1}{n}+\dfrac{1}{m} \ge p \right)$$

Multi-variable limits can often be tricky to figure out precisely what they converge to (or if they even converge at all). Luckily, in this instance, finding the value of the limit is easy, since the function is separable:

$$\displaystyle \displaystyle \lim_{n \to \infty}\left( \dfrac{1}{n} \right) + \lim_{m \to \infty}\left( \dfrac{1}{m} \right) \ge p$$

Why don't you try playing around with this a bit, and see if you can figure out why these two limits ultimately give a contradiction. At the end, you should end up with a conclusion that violates the original assumption that p > 0

#### Alfredo Dawlabany

##### New member
Absolutely. In fact, that's essentially what the proof does. Whether you call the number you pick p or $$\displaystyle \epsilon$$ is of no consequence. If we begin by assuming that there exists some p > 0 that is a lower bound of E, what we're really saying is that:

$$\displaystyle \forall n \in \mathbb{N}^* \: \forall m \in \mathbb{N}^* \: \left( \dfrac{1}{n}+\dfrac{1}{m} \ge p \right)$$

We can then reformulate this in terms of limits:

$$\displaystyle \displaystyle \lim_{(n,m) \to (\infty, \infty)}\left( \dfrac{1}{n}+\dfrac{1}{m} \ge p \right)$$

Multi-variable limits can often be tricky to figure out precisely what they converge to (or if they even converge at all). Luckily, in this instance, finding the value of the limit is easy, since the function is separable:

$$\displaystyle \displaystyle \lim_{n \to \infty}\left( \dfrac{1}{n} \right) + \lim_{m \to \infty}\left( \dfrac{1}{m} \right) \ge p$$

Why don't you try playing around with this a bit, and see if you can figure out why these two limits ultimately give a contradiction. At the end, you should end up with a conclusion that violates the original assumption that p > 0
Thank you

#### ksdhart2

##### Senior Member
Thank you
No problem. Although, I should note one correction I thought of after I posted, regarding the reformulation into limits bit. That only works because we know that both 1/n and 1/m are monotonically decreasing (i.e. for every $$\displaystyle n \in \mathbb{N}^*$$, we have $$\displaystyle \dfrac{1}{n+1} < \dfrac{1}{n}$$). Without this information, evaluating the limit wouldn't guarantee that the property holds for every n.

As an example of what I mean, suppose we wanted to show that $$\displaystyle \forall x \in [0, 2pi] \: \left( sin(x) \le \dfrac{1}{2} \right)$$. Here, all the limit(s) would tell us is that $$\displaystyle sin(2\pi) = 0 \le \dfrac{1}{2}$$ or $$\displaystyle sin(0) = 0 \le \dfrac{1}{2}$$, but it doesn't give us any information on any points in between. And, of course, we know that sin(x) does go above 1/2. Namely $$\displaystyle sin \left( \dfrac{\pi}{2} \right) = 1 > \dfrac{1}{2}$$, so the claim is invalidated, even though the limits indicated it should be true.