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Aug
27
comment Proving lack of convergence in $\sup$-norm
Ah, thank you Ilya
Aug
27
accepted Checking Uniform Convergence
Aug
27
asked Proving lack of convergence in $\sup$-norm
Aug
26
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
Aug
26
accepted Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$
Aug
26
asked Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$
Aug
25
comment Problem books in higher mathematics
Abstract Algebra: amazon.com/Abstract-Algebra-Manual-Problems-Solutions/dp/…
Aug
25
answered Problem books in higher mathematics
Aug
24
accepted Proving pointwise convergence to a “Dirichlet-like” function
Aug
24
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
Aug
24
revised Proving pointwise convergence to a “Dirichlet-like” function
added 112 characters in body; edited title
Aug
24
revised Proving pointwise convergence to a “Dirichlet-like” function
added 112 characters in body; edited title
Aug
24
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.
Aug
23
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?
Aug
23
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?
Aug
23
asked Proving pointwise convergence to a “Dirichlet-like” function
Aug
23
accepted Sequence of $C^1[0,1]$ functions $(f_n) \to f$ but $f \notin C^1[0,1]$
Aug
23
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?
Aug
23
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.
Aug
23
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 $x = 0$.