$f_n(x)=x^n$ not uniformly convergent on $(0,1)$
I know there are multiple posts on this same question but please verify if my argument is correct or not::
I am convinced that $$f_n \text{ is not uniformly cauchy on D if}$$
$$\exists \epsilon_0>0 \ni\;\forall n \in N \ \; \exists x_0 \in D \; \exists m_0, n_0 \geq n \ni \;$$ (Here, my $x_0, \; n_0,\; m_0$ will depend on $n$)
$$|f_{n_0}(x_0) - f_{m_0}(x_0)|\geq \epsilon_0 -->>(1)$$
Is above intepretation of not unifrom cauchy correct?
so,
$$\exists \epsilon_0=\frac18 >0 \ni\;\forall n \in N \; \exists \frac1{2^{\frac1n}} \in [0,1] \; \exists n, 2n \geq n \ni \;$$
$$|\big(\frac1{2^{\frac1n}}\big)^n - \big(\frac1{2^{\frac1n}}\big)^{2n}| = |\frac12 - \frac14| = \frac14\geq\epsilon_0$$
So $x^n$ is not uniformly cauchy and hence not uniformly convergent.
Is this correct?
More specifically, is $(1)$ correct?