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Let $*:[0,1]\times[0,1]\to[0,1]$ be a continuous $t$-norm i.e.

a) $*$ is continuous,

b) $*$ is commutative and associative,

c) $1*a=a~\forall~a\in[0,1],$

d) $a\le b,c\le d\implies a*c\le b*d.$

How to show that given $\epsilon\in(0,1)~\exists~\delta\in(0,1)$ such that $\delta*\delta*\delta<\epsilon?$

Please help.

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  • $\begingroup$ Choose $\delta < \epsilon < 1$ then $(\delta*\delta)*\delta \le (\delta*\delta)*1 =1*(\delta*\delta)=\delta*\delta \le \delta*1 = 1*\delta = \delta < \epsilon$. $\endgroup$ – fleablood Mar 31 '19 at 16:37
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You can always choose some $\delta>0$ such that $\delta < \epsilon$. We also know that $\delta < 1$. Hence, according to your composition rule (d) of $*$, we get $\delta * \delta< \epsilon * 1=\epsilon$. Repeat this to get the required result.

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More generally, to show that, for any $1 > e > 0$, for any integer $n \ge 2$, ther is a $ d > 0$ such that $ d^n < e$.

Note: $d$ and $e$ are easier to enter than $\delta$ and $\epsilon$.

Let $c = \frac1{e}-1$, so that $e = \frac1{1+c}$. Then $c > 0$ since $0 < e < 1$.

By Bernoulli's inequality, if $n \ge 2$, $(1+\frac{c}{n})^n \gt 1+c$, so $e =\frac1{1+c} \gt \frac1{(1+\frac{c}{n})^n} = (\frac1{1+\frac{c}{n}})^n $.

Therefore $\frac1{1+\frac{c}{n}}$ will work.

Note. To prove Bernoulli's inequality in the form if $x > 0$ and $n \ge 2$ then $(1+x)^n > 1+nx$.

For $n=2$, $(1+x)^2 =1+2x+x^2 \gt 1+2x$.

If true for $n \ge 2$, then

$\begin{array}\\ (1+x)^{n+1} &=(1+x)(1+x)^n\\ &>(1+x)(1+nx)\\ &=1+(n+1)x+nx^2\\ &>1+(n+1)x\\ \end{array} $

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Just to be clear: If we define $k^n:= \underbrace{k*k*...*}_{k\text{ times}}$ it's fairly straight forward to prove by induction that if $k \in [0,1]$ and if:

$k^n \le k \le 1$ (which we know is true for $n = 1$)

Then as $k^n \le k^n$ and $k\le 1$ < $k^{n+1} = k^n*k\le k^n*1=1*k^n = k^n\le k \le 1$.

So $1 \ge k \ge k^2 \ge k^3 \ge.....$.

This reduces the problem to the trivial: For any $0< \epsilon < 1$ we can find $0 < \delta < \epsilon$ and $\delta*\delta*\delta \le \delta < \epsilon$.

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