Convergence of $\max$ and $\min$ sequences of first $n$ terms of a convergent sequence

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!

• You need to test if $y_n$ is monotonic, and if it is bounded. (Then repeat for the case of $z_n$). Nov 7 '20 at 1:14
• yeah i got it .thanks Nov 7 '20 at 1:21
• What about the sequence $+1,-1,0,0,0,0,\ldots \text{ ?} \qquad$ Nov 7 '20 at 1:24
• @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. Nov 7 '20 at 1:32
• @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. Nov 7 '20 at 1:45

$$(y_n)$$ 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 $$z \le z_R < y_R \le y .$$