# Zvpunry

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 Aug27 accepted Checking Uniform Convergence Aug27 asked Proving lack of convergence in $\sup$-norm Aug26 comment Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$ I see. So since $(f_n)$ is convergent, I have that for $\epsilon >0$ there is an $N$, respectively an $M$, so that for $n \ge N$, $m \ge M$, we have $\|f_n -f \|_{\infty} < {\epsilon \over 2}$ and $\|f_m -f \|_{\infty} < {\epsilon \over 2}$. Therefore we have $\|f_n - f_m\|_{\infty} < \epsilon$. Thank you for this succinct explanation Aug26 accepted Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$ Aug26 asked Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$ Aug25 comment Problem books in higher mathematics Abstract Algebra: amazon.com/Abstract-Algebra-Manual-Problems-Solutions/dp/… Aug25 answered Problem books in higher mathematics Aug24 accepted Proving pointwise convergence to a “Dirichlet-like” function Aug24 comment Proving pointwise convergence to a “Dirichlet-like” function @BenjaLim I think that is the point as to why it does not converge uniformly, but we can always take $n$ sufficiently large as to make the difference less than $\epsilon$. We list up to $n$ simply by AC I believe Aug24 revised Proving pointwise convergence to a “Dirichlet-like” function added 112 characters in body; edited title Aug24 revised Proving pointwise convergence to a “Dirichlet-like” function added 112 characters in body; edited title Aug24 comment Proving pointwise convergence to a “Dirichlet-like” function I am not sure what you mean. Can you elaborate further on your hint? I thought I had answered your questions. Aug23 comment Proving pointwise convergence to a “Dirichlet-like” function Building off of this, would I suppose to the contrary that there is a sufficiently large $N_0$ so that there is an $x \in [0,1]$ for which $f_{N_0}(x) = 0, f(x) = 1$. Taking $n \ge N_0$, I have that for all $\epsilon > 0$, $|f_n(x) - f(x)| \le \epsilon. This can be done as many times as necessary to achieve arbitrary precision. Is this the lines along which you would approach? Aug23 comment Proving pointwise convergence to a “Dirichlet-like” function Do I appeal to the countability of$\Bbb Q$and show that$|\Bbb Q| = |\Bbb N|$implies pointwise convergence? Aug23 asked Proving pointwise convergence to a “Dirichlet-like” function Aug23 accepted Sequence of$C^1[0,1]$functions$(f_n) \to f$but$f \notin C^1[0,1]$Aug23 comment Sequence of$C^1[0,1]$functions$(f_n) \to f$but$f \notin C^1[0,1]$At least that's what I believe I'm looking for. The "all" would be replaced by "any" if the question was looking for nowhere differentiable right? Aug23 comment Sequence of$C^1[0,1]$functions$(f_n) \to f$but$f \notin C^1[0,1]$@Ahriman I'm looking for at least one point. Aug23 comment Sequence of$C^1[0,1]$functions$(f_n) \to f$but$f \notin C^1[0,1]$@timur I think that is what I mean by "cusps". For example,$|x|$has a cusp at$x = 0$. Aug23 asked Sequence of$C^1[0,1]$functions$(f_n) \to f$but$f \notin C^1[0,1]\$