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I Proving equivalence between statements about a sequence

  1. Feb 12, 2017 #1
    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.
     
  2. jcsd
  3. Feb 13, 2017 #2
    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)##!
     
  4. Feb 13, 2017 #3
    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
     
  5. Feb 13, 2017 #4

    Svein

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    • 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: |xn-L|<ε)
     
  6. Feb 13, 2017 #5
    Svein, thanks for additional information
     
  7. Feb 14, 2017 #6
    So only the second statement is equivalent of ##x_n## being bounded. So I am going to present the proof here. First lets 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.
     
  8. Feb 14, 2017 #7

    Svein

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    In your first post you said:
    Your conditions specify that L is finite and that the xn are finite for all n<N (N is arbitrary).

    Your second statement implies that from a certain n0, |xn-L|<ε (which is finite) and therefore |xn|<|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)!
     
  9. Feb 14, 2017 #8
    Yes Svein, I think ##L## is finite. May be I should explicitly say it. Other than that, is the proof I presented OK ?
     
  10. Feb 14, 2017 #9
    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.
     
  11. Feb 14, 2017 #10
    Thanks Q.B. , this chain of quantifiers can be confusing sometimes
     
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