Skip to main content
5 votes

Does a continuous surjective map on topological spaces always have a continuous right-inverse?

This is very, very far from true, for quite interesting reasons. Here are some suggestive examples. Let $f : Y \to X$ be a covering map between reasonable path-connected spaces, classified by a ...
Qiaochu Yuan's user avatar
3 votes

$f,g \in [0,1] , f<g $ , when is $U:=\{h \in C[0,1]:f(t)<h(t)<g(t), \forall t \in [0,1] \}$ a ball in $C[0,1]$ with respect to the sup metric

The credit still goes to Crostul's answer, but having a more detailed answer can't hurt, especially given that this same question has been asked again. Assume that $U$ is an open ball $B(F,r)$. Let's ...
Bruno B's user avatar
  • 5,904
2 votes

$\displaystyle \lim_{n \rightarrow +\infty} \dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right) = f(0)$

Assume $f(0)=0$. Let $\epsilon >0$. Choose $j$ such that $|f(x)|<\epsilon$ for $ x <\frac 1j$. Now split $\dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right)$ into sum ...
geetha290krm's user avatar
  • 39.2k
2 votes

$\displaystyle \lim_{n \rightarrow +\infty} \dfrac{1}{\log(n)}\sum_{k=1}^n \dfrac{1}{k}f\left(\dfrac{k}{n} \right) = f(0)$

Here I show that For any bounded function function $f$ on $[0,1]$ that is continuous at $0$ $$\frac{1}{\log n}\sum^n_{k=1}\frac1k f(\frac{k}{n})\xrightarrow{n\rightarrow\infty}f(0)$$ Recall that ...
Mittens's user avatar
  • 40.8k
2 votes

Proof that a function which is continuous on $[a,b]$ is uniformly continuous (Spivak)

First of all, yes, the conclusion should be that the lemma allows us to say that $f$ is $\epsilon$-good on $[a, \alpha_\epsilon+\delta_0]$, which then contradicts the fact that $\alpha_\epsilon$ is ...
5xum's user avatar
  • 124k

Only top scored, non community-wiki answers of a minimum length are eligible