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

Please help. 
 A: 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}
$
A: 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 $\delta*\delta*\delta \le \delta < \epsilon$.
A: 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. 
