15m

asked  Surniversal spaces 
11h

comment 
Snakelike continua and universal images
My general comment addressed to the whole MO: let's make MO posts more selfcontained, let them include explicit definitions. 
2d

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Can you write $\mathbb R^2$ as a disjoint union of two totally disconnected sets?
Dear @LasseRempeGillen, it's Janiszewski. Yes, I'll try to find your answer to the q. referenced by Kristal. 
Apr
24 
comment 
Applications of space filling curves
I applied, during the first half of 1985, the Hilbert curve to the image compression. 
Apr
23 
comment 
is there any more odd nearperfect number?
I am enthusiastic about almost nearly quasi subperfect pseudointegers. 
Apr
23 
comment 
is there any more odd nearperfect number?
=== Definition? === 
Apr
23 
answered  What is the correct term for “cocovering” designs 
Apr
18 
comment 
Do Peano curves provide a counterargument to Grothendieck's critique?
"...regarded as exhibiting the naturality of phenomena..." ? 
Apr
9 
comment 
Are there only finitely many nearperfect numbers with more than 4 distinct prime divisors?
I don't see any argument against the value of the q.f.'s question. Thus, I'd like to keep it open. 
Mar
24 
comment 
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
@HJRW  thank you for correcting me. My vague recollection was wrong in two directions. Let me repeat after you, yes, it was Markov and 4dimensional manifolds. 
Mar
23 
comment 
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
@SebastianGoette  actually, I applied Wall's homotopic example to obtain analogous example for compact spaces in the shape theory, it was simple (however, two guys had stolen my result in a broad daylight; I guess they needed their tenure bad or something like this). 
Mar
23 
comment 
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
@AlexDegtyarev, I remember an old similar result by a Soviet mathematician (I think that it was Malcev, the father) for 3dimensional manifolds. 
Mar
23 
comment 
No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
The proper (sotospeak) version of the problem is: *** there doesn't exist any algorithm which decides about two spaces whether or not they are homeomorphic. *** Here we can actually restrict our attention to finite polyhedra only. 
Mar
14 
comment 
Primary structures in $\mathbb Q$
Emil, thank you for your answer. And let's still hope for more. 
Mar
14 
revised 
Primary structures in $\mathbb Q$
cosmetic 
Mar
14 
revised 
Primary structures in $\mathbb Q$
accidental word omission corrected 
Mar
8 
comment 
If all balls around fixed basepoints are isometric, are the spaces as well (length spaces)?
I feel that it would not hurt to add a definition (rather than a link) of "proper length". 
Mar
2 
revised 
Does elementwise concavity guarantee joint concavity?
Further possibilities. (Earlier my additional statement got accidentally erased). 
Mar
2 
revised 
Does elementwise concavity guarantee joint concavity?
Further possibilities. 
Mar
2 
answered  Does elementwise concavity guarantee joint concavity? 