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

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 $$\delta*\delta*\delta<\epsilon?$$

• Choose $\delta < \epsilon < 1$ then $(\delta*\delta)*\delta \le (\delta*\delta)*1 =1*(\delta*\delta)=\delta*\delta \le \delta*1 = 1*\delta = \delta < \epsilon$. – fleablood Mar 31 '19 at 16:37

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.

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

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

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 $$\delta*\delta*\delta \le \delta < \epsilon$$.