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May
31
answered Is the box topology good for anything?
May
31
comment Totally disconnected implies base of closed sets?
Sure. +1 for a nice answer.
May
31
comment Continuous images of compact sets are compact
The proof looks good.
May
30
awarded  Vox Populi
May
30
revised Hausdorff dimension of Cantor set
added 113 characters in body
May
30
revised Hausdorff dimension of Cantor set
added 1 characters in body
May
30
answered Hausdorff dimension of Cantor set
May
30
revised $\sigma$-field from $(X,Y)$ and Probability measure comparison
Added set brackets where necessary.
May
30
awarded  Suffrage
May
30
suggested approved edit on $\sigma$-field from $(X,Y)$ and Probability measure comparison
May
30
comment $\sigma$-field from $(X,Y)$ and Probability measure comparison
In that case, the sets are correct. Btw notice that while inside dollars you have to type \{ to make the regular { symbol appear. I edited the set brackets accordingly, if you want you can review it after it's visible.
May
30
comment $\sigma$-field from $(X,Y)$ and Probability measure comparison
Note that in your attempt the sets $\sigma(X)$ and $\sigma(Y)$ are not $\sigma$-fields to start with.
May
30
revised Equivalent definition of absolutely continuous
added spacing in LaTeX for easier readability.
May
30
suggested approved edit on Equivalent definition of absolutely continuous
May
30
revised $\sigma$-algebras of $\{1,2,3\}$
Added latex symboling.
May
30
suggested approved edit on $\sigma$-algebras of $\{1,2,3\}$
May
29
answered $\sigma$-algebras of $\{1,2,3\}$
May
29
comment $\sigma$-algebras of $\{1,2,3\}$
The given example is actually closed under complements; so this is not where it fails to be a $\sigma$-algebra.
May
29
revised Compactness properties imply continuity
deleted 4 characters in body
May
29
comment Compactness properties imply continuity
Thanks @MTurgeon, I updated my answer:)