##(a_n) ## has +10,-10 as partial limits. Then 0 is also a partial limit

[CGandC]

Problem: If sequence $(a_n)$ has $10-10$ as partial limits and in addition $\forall n \in \mathbb{N}.|a_{n+1} − a_{n} |≤ \frac{1}{n}$, then 0 is a partial limit of $(a_n)$.

Proof : Suppose that $0$ isn't a partial limit of $(a_n)$. Then there exists $\epsilon_0 > 0$ and $N \in \mathbb{N}$ s.t. $\forall n \geq N. | a_n - 0 | \geq \epsilon_0$ ( the defintion for a partial limit $L$ of some sequence sequence $(b_n)$ is $\forall \epsilon>0. \forall N \in \mathbb{N} .\exists n \geq N. | b_n - L | < \epsilon$, so we just used the negation of it in our proof ).
Hence $\forall n \geq N. a_n \leq -\epsilon_0 \lor a_n \geq \epsilon_0$. There exists $N^* \in \mathbb{N}$ s.t. $N^* \geq max \{ N,\frac{1}{2\epsilon_0} \}$.
Instantiatng $N^*$ we get that $a_{N^*} \leq -\epsilon_0 \lor a_{N^*} \geq \epsilon_0$. Suppose without lose of generality that $a_{N^*} \leq -\epsilon_0$.
We define the set $A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \epsilon_0 \}$, since $+10$ is a partial limit of $(a_n)$ then $A \neq \emptyset$ , Hence $A$ has a minimum ( since $A$ is a set of naturals, it has minimum [ we use well-ordering principle here ] ) which we designate as $M = min A$.
Hence $a_M \geq \epsilon_0$ and $a_{M-1} \leq -\epsilon_0$, Hence $a_M - a_{M-1} \geq 2 \epsilon_0 > \frac{1}{N^*} \geq \frac{1}{M-1}$ ( Note: since we have $M > N^*$ then $M -1 \geq N^*$ hence $\frac{1}{N^*} \geq \frac{1}{M -1}$ ), so $|a_M - a_{M-1} | \geq 2 \epsilon_0 > \frac{1}{N^*} \geq \frac{1}{M-1}$. But after instantiation ( of $M-1$ in $\forall n \in \mathbb{N}.|a_{n+1} − a_{n} |≤ \frac{1}{n}$ ) we get $|a_M - a_{M-1} | \leq \frac{1}{M-1}$ which is a contradiction.

My question: I didn't fully understand why the set $A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \epsilon_0 \}$ is non-empty, how does the fact that $+10$ is a partial limit help me understand this? ( since $+10$ is a partial limit we have $\forall \epsilon>0. \forall N \in \mathbb{N} .\exists n \geq N. | a_n - 10 | < \epsilon$ , but I have no idea how instantiating here with $\epsilon_0$ or $N^*$ helps me)

 

[Office_Shredder]

As long as, let's say, $\epsilon < 5$, you get by the triangle inequality that $|a_n-10|<\epsilon$ implies $a_n>\epsilon$.

The thing you want is obviously not true if $\epsilon_0$ is too large, for example if $\epsilon_0=100$ you could easily construct a sequence that never has a term larger than this.

 

[CGandC]

Can you please explain how did you arrive from $| a_n - 10 | < \epsilon$ to $a_n > \epsilon$ ? by triangle inequality I get $( a_n - 10 ) \geq | a_n - 10 | < \epsilon \iff a_n < 10 + \epsilon$.

 

[Office_Shredder]

Sorry, it's not even the triangle inequality, I don't know why I wrote that.

$|a_n-10|<\epsilon$ implies $a_n>10-\epsilon$. If $\epsilon<5$, then $10-\epsilon >5 >\epsilon$.

 

[CGandC]

But I don't know what $\epsilon_0$ is, I just know it exists. So what would ensure to me that $\epsilon > \epsilon_0$ ?

 

[Office_Shredder]

I think it might help to take a step back and think about the broad strokes of what this proof is trying to do.

Since $10$ and $-10$ are partial limits, you could have a sequence that looks like $10,-10,10,-10$. But we have this condition that $|a_n-a_{n+1}|<1/n$. So instead of skipping between $-10$ and $10$, it hsa to slowly move between them. Once you get to $n=1000$, say, your sequence kind of has to move back and forth like $-10,-9.999,-9.998,...,-.001,0,.001,002,...,9.998,9.999,10$. And since you can only take very tiny steps, one of those steps has to end up being close to 0.

I feel like the best thing to do would be to restructure the proof a bit to more naturally fit this high level description. Are you allowed to do that, or are you really required to just understand why that red line is true? For example if $\epsilon_0$ is a really large number, that set might be false, and you kind of have to make some other arguments for why $\epsilon_0$ can't be a really large number to begin with.

 

[CGandC]

Well, there is a constructive proof for the theorem above which does not go by means of contradiction, but I understood it and did not write it here. The proof I've presented above is the one which goes means of contradiction and the red text is the part I don't undestand. I can restructure it but I'd like it still to be under the initial assumption that $0$ is not a partial limit of $(a_n)$. Do you have a way to restructure it so it can accommodate all cases of $\epsilon_0$ ? ( whether it is really big number or really small number ).

 

[CGandC]

$|a_n-10|<\epsilon$ implies $a_n>10-\epsilon$. If $\epsilon<5$, then $10-\epsilon >5 >\epsilon$.

I think I understand what you are trying to say:
We'll want to proof to work for example when I'll choose $\epsilon = 1$. ( then we'd have $5 > \epsilon = 1$ ) So we'll choose $\hat{\varepsilon_0}:=\min\{ 1, \varepsilon_0 \}$. Then clearly this quantity is still smaller than $|a_n|$ and it is smaller or equal than 1.
And then I could reformulate the proof for $\hat{\varepsilon_0}$ and so I could define the set as $A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \hat{\varepsilon_0} \}$ instead of just with $\epsilon_0$. And then it will surely be satisfied that $a_n = 10 - (10-a_n)\geq 10 -\vert 10-a_n\vert \geq 10-\varepsilon > \hat{ \varepsilon_0 }$ , for $\epsilon = 1$. That'll work?

 

[Office_Shredder]

I think that's right.

 

[WWGD]

Not sure what a partial limit is, is it a limit point? Then you can have $10(-1)^n$ without 0 being a limit point.

 

[Office_Shredder]

Yes, a partial limit is just a limit point of the sequence, and check the op again. It assumes the sequence is cauchy in a specific way.

 

[Infrared]

It assumes the sequence is cauchy in a specific way.

Actually the condition $|a_{n+1}-a_n|\leq 1/n$ does not imply that the sequence is Cauchy, even with the added condition that $\pm 10$ are subsequential limits (and in fact it cannot be Cauchy for the problem statement to hold). Since the harmonic series diverges, there is "enough room" for the sequence to to go back and forth between those two limits.

 

[Office_Shredder]

Oh sorry you're right.

 

[CGandC]

Restructured Proof : Suppose that $0$ isn't a partial limit of $(a_n)$. Then there exists $\epsilon_0 > 0$ and $N \in \mathbb{N}$ s.t. $\forall n \geq N. | a_n - 0 | \geq \epsilon_0$ ( the defintion for a partial limit $L$ of some sequence sequence $(b_n)$ is $\forall \epsilon>0. \forall N \in \mathbb{N} .\exists n \geq N. | b_n - L | < \epsilon$, so we just used the negation of it in our proof ).
Hence $\forall n \geq N. a_n \leq -\epsilon_0 \lor a_n \geq \epsilon_0$.
Define $\hat{\varepsilon_0}:=\min\{ 1, \varepsilon_0 \}$ ( Notice that $\hat{\varepsilon_0} \leq 1 , \varepsilon_0 \iff - \hat{\varepsilon_0} \geq -1 , - \varepsilon_0$ ). Hence$\forall n \geq N. a_n \leq -\epsilon_0 \lor a_n \geq \epsilon_0 \iff \forall n \geq N. a_n \leq -\hat{\varepsilon_0} \lor a_n \geq \hat{\varepsilon_0}$
There exists $N^* \in \mathbb{N}$ s.t. $N^* \geq max \{ N,\frac{1}{\hat{\varepsilon_0}} \}$.
Instantiating $N^*$ we get that $a_{N^*} \leq -\hat{\varepsilon_0} \lor a_{N^*} \geq \hat{\varepsilon_0}$. Suppose without lose of generality that $a_{N^*} \leq - \hat{\varepsilon_0}$.
We define the set $A = \{ n \in \mathbb{N} : n>N^* , a_n \geq \hat{\varepsilon_0} \}$, since $+10$ is a partial limit of $(a_n)$ then $A \neq \emptyset$ ( " +10 is a partial limit " means $\forall \epsilon>0 \forall N \in \mathbb{N} . \exists n > N. | a_n - 10 | < \epsilon$ , hence if we instantiate $\epsilon = 5 , N = N^*$ we get that there exists $n> N^*$ s.t. $\hat{\varepsilon_0} < 5 = 10-\epsilon < a_n < 10 + \epsilon$ ) , Hence $A$ has a minimum ( since $A$ is a set of naturals, it has minimum [ we use well-ordering principle here ] ) which we designate as $M = min A$.
Hence $a_M \geq \hat{\varepsilon_0}$ and $a_{M-1} \leq -\hat{\varepsilon_0}$ ( why $a_{M-1} \leq -\hat{\varepsilon_0}$? since we know it can't be the case that $a_{M-1} \geq \hat{\varepsilon_0}$, then by instantiating $M-1$ in $\forall n \geq N. a_n \leq -\hat{\varepsilon_0} \lor a_n \geq \hat{\varepsilon_0}$ we get $a_{M-1} \leq -\hat{\varepsilon_0}$ ), Hence $a_M - a_{M-1} \geq 2 \hat{\varepsilon_0} > \frac{1}{N^*} \geq \frac{1}{M-1}$ ( Note: since we have $M > N^*$ then $M -1 \geq N^*$ hence $\frac{1}{N^*} \geq \frac{1}{M -1}$ ), so $|a_M - a_{M-1} | \geq 2 \hat{\varepsilon_0} > \frac{1}{N^*} \geq \frac{1}{M-1}$. But after instantiation ( of $M-1$ in $\forall n \in \mathbb{N}.|a_{n+1} − a_{n} |≤ \frac{1}{n}$ ) we get $|a_M - a_{M-1} | \leq \frac{1}{M-1}$ which is a contradiction ( to the fact that $|a_M - a_{M-1} | > \frac{1}{M-1}$ ).