Uniform continuity choosing delta problem

Question

I am trying to show that $x^2$ is uniformly continuous on the set union of intervals [n,n+$n^{-2}$] for all positive integer n. I have also checked the previous posts. But my question is that for a given epsilon we need to choose $\delta$ smaller than $\frac{\epsilon}{2(n+n^{-2})}$. This will ensure uniform continuity. Here delta depends on interval where two elements are. for n=5 delta changes or for n=6 also another delta. In uniform continuity for every epsilon there exists a delta which satisfies for all points in delta distance. Here there are several deltas if I am not mistaken. Can you clarify this issue? I am bit confused thanks a lot.

Answer 1

Suppose that you have a function $f\colon[-1,1]\longrightarrow\Bbb R$ and consider the problem: is the restriction of $f$ to $[-1,0]$ continuous? Asserting that it is means that, for any $\varepsilon>0$, there is some $\delta_1>0$ such that$$(\forall x,y\in[-1,0]):|x-y|<\delta_1\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$Now, consider the problem: is the restriction of $f$ to $[0,1]$ continuous? Asserting that it is means that, for any $\varepsilon>0$, there is some $\delta_2>0$ such that$$(\forall x,y\in[0,1]):|x-y|<\delta_2\implies\bigl|f(x)-f(y)\bigr|<\varepsilon.$$So, you have two $\delta$'s, $\delta_1$ and $\delta_2$, which may be distinct. There is nothing wrong with that. You wanted to establis that there was one $\delta$ for each interval, and, again, it may will happen that they are not the same.

So, in your situation, on any interval $\left[n,n+\frac1{n^2}\right]$ you shall have a certain number $\delta_n$. And there's nothing wrong if there is no $\delta$ that works for all thos intervals. There is no such $\delta$ if the intervals are of the for $[n,n+1]$, but the restriction of $x^2$ to any such interval is still uniformly continuous.