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Let $(x_n)$ be a sequence of real numbers converging to $x$.

Define sequences $y_n:=\max\{x_1,x_2,\ldots,x_n\}$ and $z_n:=\min\{x_1,x_2,\ldots,x_n\}$

Now do the sequences $(y_n),(z_n)$ converge? If $(x_n)$ is monotonic , then both these sequences are convergent. What can we say in general case? Please help me with this. Thanks!

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  • $\begingroup$ You need to test if $y_n$ is monotonic, and if it is bounded. (Then repeat for the case of $z_n$). $\endgroup$
    – Michael
    Nov 7 '20 at 1:14
  • $\begingroup$ yeah i got it .thanks $\endgroup$
    – Eklavya
    Nov 7 '20 at 1:21
  • $\begingroup$ What about the sequence $+1,-1,0,0,0,0,\ldots \text{ ?} \qquad$ $\endgroup$ Nov 7 '20 at 1:24
  • $\begingroup$ @MichaelHardy: OP's question seems to be just about convergence, not necessarily to the same limit. In your example, all the sequences $(x_n),(y_n),(z_n)$ seems to converge to $0,+1,-1$ respectively. $\endgroup$ Nov 7 '20 at 1:32
  • $\begingroup$ @OP: like Michael mentioned, both $y_n$ and $z_n$ are monotone increasing and decreasing respectively since $\max A\leq\max B$ and $\min A\geq\min B$ for $A\subseteq B\subseteq\Bbb R$ and they are both bounded above by $\sup\{x_n\}$ and below by $\inf\{x_n\}$, both of which exist since $(x_n)$ is convergent, and therefore bounded. $\endgroup$ Nov 7 '20 at 1:45
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$(y_n)$ is monotically increasing. It is also bounded: There exists $N$ such that $\lvert x_n - x \rvert < 1$ for $n \ge N$. Thus $x_n < x + 1$ for $n \ge N$ and thefore $y_n \le \max(y_N,x+1)$ for all $n$. Thus $(y_n)$ converges to some $y \in \mathbb R$. Similarly $(z_n)$ converges to some $z \in \mathbb R$.

However, $z < y$ unless $(x_n)$ is constant: If $(x_n)$ is not constant, then $x_N \ne x_M$ for suitable $N, M$. Let $R = \max(N,M)$. Then $z_R < y_R$ and thus $$z \le z_R < y_R \le y .$$

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