bio | website | andrej.com |
---|---|---|
location | Ljubljana | |
age | 44 | |
visits | member for | 5 years, 7 months |
seen | 15 hours ago | |
stats | profile views | 7,503 |
I am a professional mathematician. My area of research is logic, constructive and computable mathematics, category theory, and semantics of programming languages.
May 20 |
reviewed | Close Copula theory on discrete random variables |
May 20 |
reviewed | Leave Open Which spaces admit bump functions? |
May 18 |
comment |
Problem with a proof in Wellfounded trees in categories
You could also write an e-mail to Erik Palmgren. |
May 18 |
comment |
Problem with a proof in Wellfounded trees in categories
This is just a guess, but maybe you need to use $W$-induction to simultaneously define $r$ and show that it is extensional. |
Apr 28 |
comment |
Is $\textbf{FHILB}$ locally regular?
How close are we to concluding that $\mathbf{FHilb}$ is regular because it's aglebraic? (It isn't quite algebraic, but seems "mostly" so.) |
Apr 18 |
revised |
Minimal totally separated spaces
editing just so I can cancel my downvote |
Apr 18 |
comment |
Minimal totally separated spaces
It does not matter. There are Hausdorff spaces which match your definition ("clopens separate"), but they are not $0$-dimensional in the standard sense (i.e., small inductive dimension). Is there a standard notion of dimension for which your definitions means "$0$-dimensional"? |
Apr 18 |
comment |
Minimal totally separated spaces
Ah, excellent. And so your last remark also shows that we do not even have to do anything with Stone spaces to find a counter-example: just take a totally disconnected space which is not $0$-dimensional. |
Apr 18 |
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Minimal totally separated spaces
Oh I see, you're saying you'll find an example which is not only totally disconnected (what OP called "0d") but is also zero-dimensional (clopens form a basis). |
Apr 18 |
comment |
Minimal totally separated spaces
For small inductive dimension, zero-dimensionality is equivalent to "clopens form a basis" (see Wikipedia if you trust it, en.wikipedia.org/wiki/Zero-dimensional_space). A space is totally disconnected iff its components are singletons, see en.wikipedia.org/wiki/Totally_disconnected_space. This is equivalent to saying that every two points can be separated by a clopen, which is what you wrote. |
Apr 18 |
comment |
Minimal totally separated spaces
Why are you allowed to make the assumption that the topology of $X$ is generated by the clopen sets? |
Apr 18 |
comment |
Minimal totally separated spaces
Your definition of "zero-dimensional" is actually known as totally disconnected. Zero-dimensionality is usually defined as "has a basis consisting of clopen sets". |
Apr 13 |
awarded | Nice Answer |
Apr 11 |
comment |
Axiom of choice for sets of finite sets
In the formulation of $AC(Z)$ you can drop the condition $x \neq \emptyset$ as it isn't doing anything. For the empty family you always have a choice function. |
Apr 1 |
comment |
Mathematicians wearing hats on arbitrary total orders
Beautiful, and with a nicely hidden use of choice in the second sentence of the proof ;-) |
Mar 24 |
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About the proof of the proposition “there exists irrational numbers a, b such that a^b is rational”
It is sad that this is a research-level question. |
Mar 24 |
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About the proof of the proposition “there exists irrational numbers a, b such that a^b is rational”
I just feel like saying $(\sqrt{2})^{\log_2 9}$. But more seriously, it is unpedagocial to speak about what we know and do not know when explaining the form of a given proof. The proof in question is not constructive because it employs the inference rule of excluded middle -- and the proof remains non-constructive no matter what else you prove or know. However, the proof can be transformed to a different proof which is constructive (assuming there is a constructive proof of irrationality of $\sqrt{2}^{\sqrt{2}}$). |
Mar 12 |
awarded | Good Answer |
Mar 12 |
revised |
What does a theoretical mathematician do?
Removing gender pronouns because someone complained about them. |
Mar 9 |
comment |
A topological concept dual to compactness
Hmm, @PaulTaylor will come to the rescue here. Maybe I am mixing up external and internal definitions. In his setting we actually work internally with exponentials, so we require an "internal" right adjoint to $\Sigma^{!} : \Sigma^1 \to \Sigma^X$. This makes a difference because the said right adjoint needs to be continuous as well. Is that the problem? |