Proving equivalence between statements about a sequence

[issacnewton]

Hello

Let $(x_n)_{n=1}^\infty$ be a real sequence and $L \in \mathbb{R}$. Consider the following conditions on
$(x_n)_{n=1}^\infty$ and $L$. $$\forall \varepsilon > 0,~ \forall n_0 \in \mathbb{N},~\exists n \in \mathbb{N}\mbox{ so that } (n \ge n_0 \mbox{ and } |x_n - L| < \varepsilon) \cdots\cdots(1)$$ $$\exists \varepsilon >0, \exists n_0 \in \mathbb{N},\mbox{ so that }\forall n \in \mathbb{N},(n \ge n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(2)$$ $$\exists n_0 \in \mathbb{N}\mbox{ so that }\forall \varepsilon >0, \forall n\in \mathbb{N},(n \ge n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(3)$$ Which one of the following is equivalent to $(x_n)_{n=1}^\infty$ being bounded ? $$\exists L\in\mathbb{R}\mbox{ such that (1) holds}$$ $$\exists L\in\mathbb{R}\mbox{ such that (2) holds}$$ $$\exists L\in\mathbb{R}\mbox{ such that (3) holds}$$ Now when $(x_n)_{n=1}^\infty$ is bounded, $\exists M >0$ such that $\forall n \in \mathbb{N}$ we have $|x_n| \leqslant M$. I can this why this leads to the first condition, but I am trying to prove this. I have tried negating the goal but all these quantifiers are causing lot of confusion and I am lost. Any guidance will be helpful.

 

[Q.B.]

Hi,

$(1)$ is verified by $x_n=L$ when $n$ is even and $x_n=L+n$ when $n$ is odd (indeed, for $\epsilon>0$ and $n_0$, define $n=2n_0$ which satisfies the condition) but this sequence is clearly unbounded.

$(3)$ leads to $x_n=L, \forall n \geqslant n_0$ (indeed, take given the $n_0$ a $x_n$ with a greater $n$ which would not be $L$. Then you can take $\epsilon$ to be $\frac{1}{2}\mid x_n-L \mid$ and it contradicts $(3)$).

So you're left with $(2)$!

 

[issacnewton]

Q.B. For the first statement, I thought, its equivalent to $x_n$ being bounded. I spent lot of time in vain trying to prove this. But I could not think of a counter example like you did. When I thought of divergent sequences, I thought they either increase, decrease or oscillate. It just did not occur to me that some terms of the sequence might remain constant as $n \Longrightarrow \infty$. This was a huge help. I will try to prove that $(2)$ is equivalent with $x_n$ being bounded

 

[Svein]

• Your first statement is the definition of L being a cluster point (which does not mean boundedness)
• Your second statement says that the sequence is bounded from a certain n₀
• Your third statement is a mixed-up definition of convergence (should be (∀ε>0)(∃n₀)(∀n>n₀: |x_n-L|<ε)

 

[issacnewton]

Svein, thanks for additional information

 

[issacnewton]

So only the second statement is equivalent of $x_n$ being bounded. So I am going to present the proof here. First let's consider the forward direction. Assuming that $x_n$ being bounded I have to prove that $$\exists L\in\mathbb{R}~\exists \varepsilon >0, \exists n_0 \in \mathbb{N},\mbox{ so that }\forall n \in \mathbb{N},(n \geqslant n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(2)$$ I am going to do contrapositive proof here. So I am going to assume $$\forall L\in\mathbb{R}~\forall \varepsilon >0,\forall n_0 \in \mathbb{N},~\exists n \in \mathbb{N} (n \geqslant n_0) \mbox{ and } |x_n-L| \geqslant \varepsilon \cdots\cdots(A)$$ and I will need to prove that $$\forall M>0 ~ \exists n \in\mathbb{N}~ |x_n| > M \cdots\cdots(B)$$ Let $M>0$ be arbitrary. Now in the statement $(A)$, choose $L=0$, $\varepsilon = 2M$ , and $n_0 = 1$. So we have $n_1 \in\mathbb{N}$ such that $n_1 \geqslant 1$ and $|x_{n_1}| \geqslant 2M > M$, which means that $|x_{n_1}| > M$. Since $M>0$ is arbitrary, this proves the statement $(B)$. Hence by contrapostive proof, I proved statement $(2)$ by assuming that $x_n$ is bounded. Now I will go for other direction. I will assume that $$\exists L\in\mathbb{R}~\exists \varepsilon >0, \exists n_0 \in \mathbb{N},\mbox{ so that }\forall n \in \mathbb{N},(n \geqslant n_0 \Longrightarrow |x_n-L| < \varepsilon)\cdots\cdots(C)$$ and I will need to prove that $x_n$ is bounded, i.e. $$\exists M>0~ \forall n \in \mathbb{N}~ |x_n| \leqslant M\cdots\cdots(D)$$ So from statement $(C)$, we have some $L\in\mathbb{R}$, some $\varepsilon >0$ and some $n_0 \in \mathbb{N}$. And I have $$\forall n\geqslant n_0 ~ |x_n-L| < \varepsilon$$ So $\forall n\geqslant n_0$, I have $$-\varepsilon-|L| \leqslant L-\varepsilon <x_n < L+\varepsilon \leqslant \varepsilon + |L|$$ which means that $\forall n\geqslant n_0$, I have $|x_n| \leqslant \varepsilon+|L|$. Its to be noted that $\varepsilon+|L| > 0$. Now define $M$ as follows $$M = \max\{1+|x_1|, 1+|x_2|,\cdots 1+ |x_{n_0-1}|, \varepsilon+|L|\}$$ Its clear from the way I have defined $M$ , that $M>0$. Also note that $\forall n < n_0$, I have $|x_n| \leqslant M$. Its also clear from the definition of $M$, that $\forall n\geqslant n_0$, it is true that $|x_n| \leqslant M$. So I proved an existence of $M>0$ such that $\forall n\in\mathbb{N}~ |x_n| \leqslant M$. Hence the statement $(D)$ is proven and hence the sequence $x_n$ is bounded once we assume $(C)$. Since both the directions are proven, this means that $x_n$ being bounded is equivalent to $$\exists L\in\mathbb{R}~\exists \varepsilon >0, \exists n_0 \in \mathbb{N},\mbox{ so that }\forall n \in \mathbb{N},(n \geqslant n_0 \Longrightarrow |x_n-L| < \varepsilon)$$ I hope my proof is correct.

 

[Svein]

In your first post you said:

Let (xn)∞n=1(x_n)_{n=1}^\infty be a real sequence and L∈RL \in \mathbb{R}. Consider the following conditions on (xn)∞n=1(x_n)_{n=1}^\infty and LL.

Your conditions specify that L is finite and that the x_n are finite for all n<N (N is arbitrary).

Your second statement implies that from a certain n₀, |x_n-L|<ε (which is finite) and therefore |x_n|<|L|+ε which is finite since L is finite.

In the same manner, the correct version of your third statement implies exactly the same thing, only here you can choose an arbitrary small ε.

The crucial point is the statement (∀n>n₀)!

 

[issacnewton]

Yes Svein, I think $L$ is finite. May be I should explicitly say it. Other than that, is the proof I presented OK ?

 

[Q.B.]

Hi, yes I think it is correct! Note that you didn't need to do a contrapositive proof for the first part, you could just build $L$ and $\epsilon$ from the $M$ you would have had (which looks very much like what you did). But it doesn't make your proof less right.

 

[issacnewton]

Thanks Q.B. , this chain of quantifiers can be confusing sometimes