I’m reading Bosch’s *Lectures on Formal and Rigid Geometry*. And I’m have some problems. Let $R$ be an adic ring. How to prove that the completion of $R$ constructed by dividing the ring of all Cauchy sequences in $R$ by the ideal of all zero sequences is Hausdorff? And consider the inverse limit $widehat{R}=lim_{n>0} R/mathfrak{a}^{n}$, where $mathfrak{a}$ is the ideal of definition of $R$, how to prove that it is Hausdorff?

# Tag: Hausdorff

## How do I see that the Hausdorff measure of slices $H^s(B)_x$ is a measurable function?

$Bsubset mathbb{R}^2$ is a Borel set. Define the slices $B_x:= {y in mathbb{R}: (x,y) in B }$.

If $lambda$ denotes the Lebesgue measure, presentations of Fubini’s theorem often include that fact that the function $lambda(B_x)$ is measurable.

**Question:** If $H^s$ denotes the $s$th Hausdorff measure, how do I see that the function $H^s(B_x)$ is measurable? $(star)$

I came across this while looking at Marstrand’s slice theorem in a book. The authors suggest to use a Monotone class argument wherein one would have to show the following

- If $B=Utimes V$ then, $H^s(B_x)= mathbb{1}_U(x)cdot H^s(V)$ which is measurable.
- If $B$ is a finite union of disjoint rectangles then again $H^s(B_x)$ is measurable.
- If $B_n$ is an increasing family of sets each of which satisfies $(star)$ then $H^sleft((cup B_n)_xright) = H^s(cup (B_n)_x) = lim H^s((B_n)_x)$ which is again measurable.
- If $B_n$ is a decreasing family of sets each of which satisfies $(star)$, one would like to show the same for $H^s((cap B_n)_x)$. However, this is equal to $H^s(cap (B_n)_x)$. This in general won’t be $lim H^s((B_n)_x)$ since we don’t know if any of the terms has finite $H^s$ measure.

I don’t see how to prove the last point in the absence of $sigma$-finiteness. Am I missing something easy?

## axiom of choice – Does ZF + BPI alone prove the equivalence between “Baire theorem for compact Hausdorff spaces” and “Rasiowa-Sikorski Lemma for Forcing Posets”?

Rasiowa-Sikorski Lemma (for forcing posets)is the statement: For any p.o. $mathbb{P}$ (i.e. $mathbb{P}$ is a reflexive transitive relation) and for any countable family of dense subsets of $mathbb{P}$ there is a generic filter which intersects all dense subsets of the countable family. It is well-known that this statement is equivalent to the Baire Category Theorem for Complete Metric Spaces – and thus it is also equivalent to the Principle of Dependent Choices.

A masters student of mine has found in the literature the following statement: “Rasiowa-Sikorski Lemma is equivalent to the Baire Category Theorem for Compact Hausdorff Spaces, modulo the Boolean Prime Ideal Theorem”. We understood this as the assertion that the theory ZF + BPI alone is able to prove the equivalence between the Baire Category Theorem for Compact Hausdorff Spaces and the Rasiowa-Sikorski Lemma.

Well, I asked my student to verify such claim, and at first glance I suggested him to follow the results 3.1 to 3.4 of Chapter II of Kunen’s book, where there are proofs for some equivalences of Martin’s Axiom at $kappa$, MA($kappa$): the idea was to discard the hypothesis “c.c.c.” and adapt the reasoning, arguing for $kappa = omega$. It turns out that it was not a good suggestion, because in 3.1 a kind of Downward-Lowenheim-Skolem argument is done, to show that it is equivalent to work with a restricted form of the forcing axiom, considering only partial orders of bounded cardinality. However, such argument seems to require the Axiom of Choice, or some part of it other than BPI.

Does any of you know if it is indeed possible to prove the equivalence between “Baire Category Theorem for Compact Hausdorff Spaces” and “Rasiowa-Sikorski Lemma for forcing posets” from ZF + BPI alone ? Any suggestions or references would be appreciated.

## real analysis – Maximal Hausdorff dimension of the set of non differentiability

Let $f, g: (0, 1) to mathbb R$ be absolutely continuous functions such that $f’ = g’$ Lebesgue almost everywhere. What is the maximal Hausdorff dimension $d$ (and corresponding Hausdorff $d$-measure) of the sets on which:

i) $f$is not differentiable?

ii) $f$ and $g$ are differentiable but with derivatives unequal?

## general topology – Find a Quotient of Hausdorff spaces not is Hausdorff

Give a example of a quotient of a Hausdorff space that is not Hausdorff.

*Attempt*

Let $X=mathbb{R}$, for $x,yin X$ we define $xsim y$ if $x=y$ or $|x|=|y|>1$.

I claim that the quotient $Xsetminus sim$ not is Hausdorff.

Let $pi: X longrightarrow Xsetminus sim$ given by $pi(x)=(x)$ and define the topology on $X setminus sim$ given by $Usubset Xsetminus sim$ is open if $pi^{-1}((U))subset X$ is open .

For $xin X$ we have two possible cases.

Case I

$xin mathbb{Q}$, namely $x=p/q$ where $(p,q)=1,qneq 0,p,qin mathbb{Z}$ WLOG $|x|>1$and then $(x)=lbrace frac{r}{s}in mathbb{Q} mid ps=qrrbrace$$bigcup$ $ lbrace -frac{r}{s}in mathbb{Q} mid ps=qrrbrace$.

Other wise $|x|<1$ then $(x)=lbrace frac{r}{s}in mathbb{Q} mid ps=qrrbrace$

Case II

$xin mathbb{I}$ then $(x)=lbrace x,-xrbrace$.

We want take use the following:

**If X is Hausdorff and $x_iin X$ for $iin 1,2,3, cdots n$ then $lbrace x_1,x_2,x_3, cdots x_nrbrace$ is closed**

Consider $(frac{1}{n})in X setminus sim $ for some $nin mathbb{N}$ fixed.

$pi^{-1}(lbrace(frac{1}{n})rbrace)=lbrace frac{k}{kn}mid kin mathbb{N} rbrace=lbrace frac{1}{n}, frac{1}{n}, cdots frac{1}{n} rbrace =lbrace frac{1}{n}rbracesubset X$ wich is closed.

Hence $lbrace(frac{1}{n})rbrace$ is closed in $Xsetminus sim$.

Now consider $bigcup_{nin mathbb{N}}lbrace left(frac{1}{n}right)rbrace in Xsetminus sim $ and notice that

$$pi^{-1}left(bigcup_{nin mathbb{N}}lbrace left(frac{1}{n}right) rbrace right)=bigcup_{nin mathbb{N}}pi^{-1}left( lbrace left(frac{1}{n}right) rbrace right)=lbracefrac{1}{n}mid nin mathbb{N} rbrace subset mathbb{R} $$

Which is clearly not closed since $0$ is limit point of the set and is not contained in it.

Hence $X setminus sim$ cannot be Hausdorff.

I miss some?

Is possible get a intuitive representation of this quotient?

## Are locally compact, Hausdorff, locally path-connected topological groups locally Euclidean?

Is every locally compact, Hausdorff, locally path-connected topological group locally Euclidean? (That would imply of course also being a Lie group). Is it true when countable basis assumption is added? I wasn’t able to find a discussion of this question in the literature on topological groups and the Hilbert 5th problem.

## general topology – Hausdorff compactly generated is locally compact?

I was trying to prove the “theorem of exponential correspondence” and “exponential law” with $X,Y,Z$ compactly generated and Hausdorff.

According to *Hedwin H.Spanier-Algebraic Topology* and Tammo tom Dieck-Algebraic topology we have that locally compact seems necessary in order to prove the statement.

The true hyphotesis given on the spaces are Hausdorff locally compact, but as far as I can see, compactly generated and Hausdorff doesn’t imply locally compact, am I right ? Are my hypothesis wrong ? Are there any counterexample.

Thanks in advance

## reference request – If $(X,tau _X)$ is a locally compact and second countable Hausdorff space, then a Borel locally finite measure is regular

Could somebody please give me one reference that contains the proof of the following theorem?

Let $(X,tau _X)$ be a locally compact and second countable Hausdorff space. Suppose that $mu _X:mathfrak{B}_Xto overline{mathbb{R}}$ is a locally finite measure in which $mathfrak{B}_X$ is the Borel $sigma$-algebra of $(X,tau _X)$. Then the following propositions are true

- For all $Binmathfrak{B}_X$ we have $mu_X(B)=inf big{mu_X (A):Bsubseteq Awedge Ain tau_X big}$
- For all $Binmathfrak{B}_X$ we have $mu_X (B)=supbig{mu_X (K):Ksubseteq Bwedge (K text{ é compact})big}$
- Given any $Bin mathfrak{B}_X$ and $varepsilon in (0,infty)$, there exists $Aintau_X$ and a closed set $F$ such that $Fsubseteq B subseteq A$ and $mu_X(Asetminus F)<varepsilon $.

I know that the above theorem is true because of the theorems that are in the book “Notes on Measure Theory” (Theorem 5.7, pg 64) and the book “Manifolds, Sheaves, and Cohomology” (Proposition 1.10, pg 4)

## lie groups – The Hausdorff dimension of $F^+_{m,n}$ singular points

Let $G:=SL(m+n,mathbb R)$ and $Gamma :=SL(m+n,mathbb Z)$ and $X:=G/Gamma$.

(1) Let $M$ denote the set of all $m times n$ matrices with real entries. A matrix $A in M$ is called $textit{singular}$ if for all $epsilon > 0$, there exists $Q_epsilon$ such that for all $Q ge Q_{epsilon}$, there exist integer vectors $p in mathbb Z^m$ and $q in mathbb Z^n$ such that

begin{equation}

|Aq+p|le epsilon Q^{-n/m} ~text{and}~

0<| q | le Q.

end{equation}

We denote the set of singular $mtimes n$ matrices by $textbf{Sing}_{m,n}$

By Dani’s correspondence principle (1985), this is equivalent to saying that $(g_t u_A mathbb Z^n)$ is divergent in the space of unimodular lattices where $g_t:=begin{bmatrix}

e^{t/m}I_m & 0 \

0 & e^{-t/n}I_n

end{bmatrix}$

and

$u_A:=begin{bmatrix}

I_m & A \

0 & I_n

end{bmatrix}$.

(2) Let $F^+:={g_t:tge 0}$ and let $D(F^+, X)$ be the set of points $X$ such that the trajectory $F^+ x$ is divergent (“leaving any compact set”).

I wonder how to show the following equation of Hausdorff dimensions:

$$dim(X)-dim(D(F^+, X))=mn- dim(textbf{Sing}_{m,n})$$

Intuitively this is very true through Dani’s correspondence (“codimension”=”codimension”!). But to prove it rigorously, are there any key theorems about the Hausdorff dimensions involved?

## gn.general topology – Regular measure in a Hausdorff space

Let $(X, beta, mu)$ be a measure space, and $(X, tau)$ be a Hausdorff topological space such that:

- $mathcal{B}(tau)subsetbeta$; where $mathcal{B} (tau)$ is the Borel set generated by $tau$.
- There is a sequence

$$ (K_n)_{nin mathbb{N}} in X $$

of compact sets such that $X$ is equal to the union of all those sets in the sequence. - The measure $mu(K)$ is finite, for all $K$ compact that is in $X$.

I want to prove that, if $mu$ is regular, then for each $Ainbeta$, and for all $epsilon>0$, there exists an open set $U supset A$ such that:

$$mu (U-A) leq epsilon$$.

To solve the problem, do we need full regularity or just outer regularity? And why does the space have to be specifically Hausdorff? Can the problem still be solved with a stronger/weaker separation axiom?

In general, what restrictions given in the problem statement can we remove/add while still having a similar solution?