Uniform convergence in open intervals

Question

I am doing Uniform convergence of functions and in this section I notice one thing that we most of the time talk about uniform convergence of sequence of functions in Closed Interval. Can we talk about this notion in an open intervals or intervals which are not closed. If it is so then help me out of this problem- Let $$f_n(x)=\frac{nx}{1+n^2x^2}.$$ This sequence of functions does not converges uniformly on $[0,1]$ as $|f_n(x)-f(x)|$ does not goes towards zero on choosing X sufficiently small close to zero say $x=1/n$. But what about in case of open interval $(0,1)$, can I use same arguement here again..... I am afraid that can I use $x=1/n$ in this case too Thanks in advance

Answer 1

In fact, since $(\forall n\in\mathbb N):\frac1n\in(0,1)$, you can use the same argument with $\frac1n$. Since$$(\forall n\in\mathbb N):f\left(\frac1n,n\right)=\frac12,$$$(f_n)_{n\in\mathbb N}$ does not converge uniformly to the null function and therefore, since it converges pointwise to that function, it doesn't converge uniformly at all.