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Suppose that $\{x_n\}$ and $\{y_n\}$ are two sequences, which converge to distinct real numbers $a$ and $b$, respectively. Need to prove that $\{x_n\}\cap\{y_n\}$ contains at most finitely many elements.

My approach:

Without loss of generality, suppose that $\left|a\right|<\left|b\right|$. Suppose also that ${z_n}:=\{x_n\}\cap\{y_n\}$ is a convergent sequence with infinitely many elements. Now, $0\leq\left|\max\{\{z_n\}\}\right|\leq \left|a\right|$, $\forall n$, thus $\exists c\in\mathbb{R}$ such that $\left| c\right| \leq \left| a\right|$, such that $\{z_n\}\to c$. Since $\{z_n\}\to c$ as $n\to \infty$, $\forall \varepsilon_1 > 0$, $\exists N_{\varepsilon_1} > 0$, such that if $n>N_{\varepsilon_1}$ then $\left|z_n-c \right|<\varepsilon_1$.

Since $\{y_n\}\to b$ as $n\to \infty$, $\forall \varepsilon_2 > 0$, $\exists N_{\varepsilon_2} > 0$, such that if $n>N_{\varepsilon_2}$ then $\left|y_n-b \right|<\varepsilon_2$. Let $\varepsilon := \max\{\varepsilon_1, \varepsilon_2\}$. Now choose $\varepsilon_1$ in such a way that for $n>N_\varepsilon$, $\left|\{x_n\}-\{y_n\}\right|<b-a+\varepsilon_1$. Now we have that $\{z_n\}\to c$ and $\{y_n\}\to b$. But this implies that, for this $n$, $\{x_n\}\cap\{y_n\}=\emptyset$, a contradiction. Thus, $z_n=\{x_n\}\cap\{y_n\}$ contains at most finitely many elements.

Please let me know if this proof is OK. Maybe it's too complicated, or has some mistakes?

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The statement $0\leq |max\{\{z_n\}\}|\leq |a|,\forall n$ is not justified and is actually not correct. First of all, by your assumption $\{z_n\}$ is convergent sequence with infinitely many terms, and it is in general not true that such a sequence has a maximum. So you should have written $0\leq |sup\{\{z_n\}\}|\leq |a|,\forall n$. Now this statement is just wrong as the following example shows

$x_n: 1,1,0,0,0,0,0\ldots$. So $x_n \to 0 = a$

$y_n: 1,1,2,2,2,2,2\ldots$. So $y_n \to 2 = b$

So $\{x_n\}\cap\{y_n\} $ is the finite sequence $\{1,1\}$

You are making this too complicated. Let $\{z_n\} = \{x_n\}\cap\{y_n\}$. Suppose that $\{z_n\}$ is an infinite sequence. Then it is an infinite subsequence of $\{x_n\}$, and as $x_n \to a \implies z_n \to a$. By the same argument $z_n\to b$. Hence $a=b$ which is a contradiction! Hence $\{z_n\}$ is a finite sequence.

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  • $\begingroup$ I have realized that a subsequence doesn't need to converge to the same limit as the "parent" sequence. So there appears to be an error in your proof. $\endgroup$ – sequence Sep 26 '16 at 6:48
  • $\begingroup$ But a subsequence always converges to the same limit as the parent sequence. Why do you think this is not true? Do you have an example? $\endgroup$ – Siddhant Sep 26 '16 at 15:08
  • $\begingroup$ For example, there is a proposition which says that $S=\{\sin(n): n\in \mathbb{Z}^+\}$ is dense in $[-1,1]$, which means that every point in $[-1,1]$ is the limit of a subsequence in $S$. $\endgroup$ – sequence Sep 27 '16 at 0:40
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    $\begingroup$ That is true, but note that the sequence $\{sin(n)\}_{n=0}^{\infty}$ is not a convergent sequence. A sequence is convergent if and only if it has exactly one limit point. $\endgroup$ – Siddhant Sep 28 '16 at 0:16
  • $\begingroup$ I see. That's very interesting. Sorry for misunderstanding. $\endgroup$ – sequence Sep 28 '16 at 1:01

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