All Questions
7 questions
18
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0
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1k
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Does there exist a continuous open map from the closed annulus to the closed disk?
(Originally from MSE, but crossposted here upon suggestion from the comments)
In this MSE post, user Moishe Kohan provides an example of a non-continuous open and closed ("clopen") function $...
- 833
0
votes
1
answer
969
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Is the pointwise supremum of a continuous function continuous?
Suppose $f(x , y)$ is continuous in both variables. For any $\epsilon > 0$ and some $y_0$, let $h_{\epsilon}(x) = \max_{y^{'}: \| y^{'} - y_0 \| \leq \epsilon} f(x , y^{'})$. It seems to me that $...
- 3
3
votes
2
answers
517
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Several definitions of approximate continuity of real functions
I found the definition of approximate continuity stated as follows:
A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ iff there exists a set $A\in \mathcal L$ such that $x_0\...
- 930
0
votes
1
answer
282
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Continuity of the Restriction Map Between Function Spaces [closed]
Let $X,Y,Z$ be Hausdorff spaces and suppose that $Z\subset X$. Endow $C(X,Y)$ and $C(Z,Y)$ with the compact-open topologies and define the map $\rho$ as
\begin{align}
\rho:&C(X,Y)\rightarrow C(Z,...
- 5,405
21
votes
3
answers
610
views
Which partitions of $[0,1]$ are collection of level sets of a real continuous function?
Let $f:[0,1]\to[0,1]$ be given. The level sets of $f$ (ie the collection of all sets of the form $\{x\in[0,1]:f(x)=y\}$, for each fixed $y\in[0,1]$) partition the domain of $f$. I am curious for set ...
- 1,014
7
votes
0
answers
227
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Uniform approximation of separately continuous functions on zero-dimensional spaces
For topological spaces $X,Y,Z$ а function $f:X\times Y\to Z$ is called separately continuous if for any $(x,y)\in X\times Y$ the restrictions of $f$ to the sets $\{x\}\times Y$ and $X\times \{y\}$ are ...
- 41.9k
7
votes
1
answer
2k
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If $S\subset\mathbb R$ is a $G_\delta$, is there a function $\mathbb R\to\mathbb R$ continuous exactly on $S$?
Let $S\subset\mathbb R$ be a $G_\delta$ set. A variation on the construction of the Thomae function (which is discontinuous on the rationals and continuous elsewhere) shows that there is a function $\...
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