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Given $f_n : X \to [0,\infty)$ define by

$f_n =\begin{cases} 1 \ \text{if }\ x\in [n,\infty) \\ 0 \ \text{if} \ x \in (-\infty,n) \end{cases}$

Here It's mention that $f_n$ is decreasing function

My question is that why $f_n$ is decreasing function ?

My thinking : $f_n$ is neither decreasing nor increasing .$f_n$ take only value $0$ and $1$ .Therefore we can said that $f_n$ is oscillating

jasmine
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1 Answers1

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I think you misunderstood the difference between an increasing/decreasing function and an increasing/decreasing sequence of functions. In the link you have attached, they say that the sequence of functions $(f_n)$ is decreasing. This is true, since for every $x\in \mathbb{R}$ one has $$f_n(x)\geq f_{n+1}(x).$$ However, the definition of an increasing/decreasing function $f$ is different. Namely for all $x<y$ one has for an increasing function $f(x)\leq f(y)$ and for a decreasing function $f(x)\geq f(y)$. For fixed $n$, your function $f_n$ is clearly increasing.

projectilemotion
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  • oks thanks u @Projectilemotion...i have some doubt take $ f_{n-1} =0$ and $f_n=1$ here $f_{n-1} \le f_n $ this show that $f_n$ is increasing but here $n$ is not fixed – jasmine Jan 09 '21 at 06:19
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    No, you have that $f_{n-1}(x)=\mathbf{1}{[n-1,\infty)}(x)$ and $f_n(x)=\mathbf{1}{[n,\infty)}(x)$. For $x< n-1$, we have $f_{n-1}(x)=f_n(x)=0$, for $x\in [n-1,n)$ we have that $1=f_{n-1}(x)\geq f_n(x)=0$. Finally for $x\geq n$ we have $f_{n-1}(x)=f_n(x)=1$. Thus in each case we have that $f_{n-1}(x)\geq f_n(x)$. – projectilemotion Jan 09 '21 at 06:25
  • okss got it now thanks u @projectilemotion – jasmine Jan 09 '21 at 06:29