## metric spaces – The decomposition theorem for the topological dimension

I'm trying to prove the decomposition theorem for the topological dimension:

This is a known result (eg Engelking) only for a non-empty separable metric space $$X$$, the small inductive dimension $$indX$$ satisfies the decomposition theorem:

$$indX leq n$$ if and only if $$X = Y cup Z$$ with $$indY leq n-1$$ and $$indZ leq0$$

We also know that in the case of separable metric spaces, the small inductive dimension and the topological dimension (also called the Lebesgue recovery dimension) are equivalent. I am given the following definition for the topological dimension:

Let $$alpha = (A_i) _ {i in I}$$ to be an open blanket of $$X$$ for some indexes $$I$$. Given $$x in X$$, to define $$ord_x ( alpha) = – 1 + # {{ in I: x in A_i$$. Now define $$ord ( alpha) = sup_ {x in X} ord_x ( alpha)$$. We say that $$beta = (B_j) _ {j in J}$$ is a finished open top finished finer than $$alpha$$ if for all $$j in J$$ it exists $$i in I$$ with $$B_j subset A_i$$. To define $$D ( alpha): = min _ { beta} ord ( beta)$$ or $$beta$$ is finer than $$alpha$$. Finally, the topological dimension of $$X$$ is given by $$dim_tX: = sup_ alpha D ( alpha),$$ or $$alpha$$ going on all the finished open covers of $$X$$.

I am now trying to prove the above theorem using the definition of the topological dimension, that is to say that $$dim_tX leq n$$ if and only if $$X = Y cup Z$$ with $$dim_tY leq n-1$$ and $$dim_tZ leq 0$$. The right-to-left direction is not hard to prove. However, the direction where we are given $$dim_tX leq n$$ and must build $$Y$$ and $$Z$$ causes me problems. I have tried the following:

Let $$alpha$$ to be an open blanket over for $$X$$ such as $$D ( alpha) = dim_tX = n$$ and $$beta$$ an open cover finished finer than $$alpha$$ with $$n = ord ( beta)$$. Consider $$I: {x in X: ord_x ( beta) = n }$$. Then I tried to do something with $$X setminus I$$ who, hopefully $$dim_t (X setminus I) leq n-1$$ and I tried to pretend that $$I$$ will be totally disconnected (or at least that I can find finer open covers of finer and finer $$beta_k$$ such as $$I _ { beta_k}: = {x in X: ord_x ( beta_k) = n }$$ has the property that $$I _ { beta_k} subset I _ { beta_ {k-1}}$$ so what $$I: = bigcap_ {k in mathbb {N}} I _ { beta_k}$$ totally disconnected), which is not very clear (and I do not know how to prove, if he is not already wrong to start). I've also thought that it might be helpful to use closed blankets $$beta$$ instead of open ones (since the definition of the topological dimension gives the same result), but again, I could not go anywhere.

In total, I'm stuck and I do not know how to prove the direction.

## Is it possible to have a Hausdorff dimension smaller than the topological dimension?

"Normal" geometric shapes have Hausdorff dimensions equal to their topological dimensions. Mandelbrot defined fractures as shapes with a Hausdorff dimension greater than their topological dimension. Is there a class of shapes with a Hausdorff dimension smaller than the topological dimension or is it impossible? If there is such a form, what are the most common examples, if not, why are they impossible.

## Topological groups with homomorphic underlying spaces, isomorphic abstract groups, and homotopic equivalent classification spaces

Define the classification space $$BG$$ of a topological topological group $$G$$ like the big realization of the nerve of $$G$$.

Let $$G$$, $$H$$ to be very topological topological groups. Assume that there is a continuous group homomorphism $$f: G rightarrow H$$ this induces an equivalence of homotopy on the underlying topological spaces. It is stated in an answer to this question that the induced card $$BG rightarrow BH$$ is an equivalence of homotopy. It is important here to have a map between groups that respects both algebra and topology simultaneously.

The question is basically what happens if we do not have such a map. Let $$G$$, $$H$$ to be very topological topological groups. Suppose their underlying abstract groups are isomorphic and their underlying topological spaces are homeomorphic. That in itself does not mean that $$G$$ and $$H$$ isomorphic as topological groups (an example in favor of choice is given by $$p$$-adical rational for different $$p$$).

But what happens if we assume in addition that the classification spaces of $$G$$ and $$H$$ are homotopy equivalent?

Note that a Notbohm theorem says that two compact Lie groups are isomorphic, because Lie groups if and only if their classification spaces are equivalent, so that this question has at least one positive answer, at least in certain contexts.

## reference query – Controllable topological topological variety with boundary = a sphere is a ball

(This question is common to Steven Karp and Thomas Lam) We must use the following fact in our document:

Theorem 1. Let $$M ^ n$$ to be a compact contractable topological variety with a limit, such as the limit $$partial M$$ is homeomorphic to a sphere $$S ^ {n-1}$$. then $$M$$ is homeomorphic to a closed balloon.

Q1. Is there a reference where this is indicated with proof? A reference we found is the theorem (slightly stronger) 10.3.3 (ii) in the Davis book. He says this for all $$n$$but only gives a sketch of proof for $$n geq 6$$.

Q2. What is the easiest way to prove Theorem 1? Here is an argument that we collected from various MO messages: The limit of $$M$$ is collared by Brown's theorem. So, we can stick a $$n$$-ball to $$M$$and by van Kampen and Mayer – Vietoris, it follows that the resulting space is a merely connected sphere of homology. So it's a sphere of Poincaré's conjecture, so $$M$$ is a ball closed by Brown's Schoenflies theorem.

Note that we do not need the interior of $$M$$ to be an open ball (and we rather demand that the limit be a sphere), which is why this question does not duplicate this question.

## General topology – Does the double space of a topological vector space always separate points?

Yes $$X$$ is a normed vector space, then the dual space $$X ^ *$$, consisting of continuous linear functions on $$X$$, separate the points. What does it mean, is that if $$x_1, x_2 in X$$, then there is a $$f in X ^ *$$ such as $$f (x_1) = 0$$ and $$f (x_2) = 1$$. This is a consequence of Hahn-Banach's theorem. But I wonder if something stronger is true.

My question is, if $$X$$ is a topological vector space, so $$X ^ *$$ necessarily separate points?

If no, does anyone know of a counter-example?

## Comparison of real topological K theory and algebraic K theory

Is there an integer $$0 with the following property:

for any commutative unital ring $$R$$, there is a Hausdorff compact space $$X$$ such as $$KO ^ i (X) about K ^ i_ {alg} (R)$$?

P.S .: I do not know if it's important, but it does not bother me particularly to formulate finiteness hypotheses (for example, $$R$$ is Noetherian).

## Comparison of topological and algebraic K theories

Let $$R$$ to be a commutative unital ring and let $$i$$ to be a non-negative integer. Is it true that there is a compact Hausdorff space $$X$$ such as $$K ^ i_ {top} (X) about K ^ i_ {alg} (R)$$?

## General topology – Two notions of delimitation in a metaphorical topological vector space

In a metric topological vector space X with the metric d, a subset A is said to be bound if it can be absorbed by any neighborhood of 0 and a subset A is said bound if its diameter relative to the metric d is finished. Delimitation always implies the delimitation d, but the opposite is not true.

I am looking for a condition for which d-boundedness implies a delimitation. In Wikipedia, in the section "Topological vector spaces", it is indicated that "the two notions of limit coincide for locally convex spaces". But there is no reference for that here. Can any one give a reference or clue to prove this statement?

## General Topology – Topological Groups Containing the Sorgenfrey Line

the Sorgenfrey line $$mathbb S$$ is the actual line with the topology generated by the base composed of all half-intervals $$[Ab)[Ab)[ab)[ab)$$ for real numbers $$a .

The Sorgenfrey line is a first accounting and non-metalisable name and is therefore not homeomorphic to a topological group.

On the other hand, the Sorgenfrey line $$mathbb S$$ is homeomorphic to a subset of a topological group. For example, the free topological group $$F ( mathbb S)$$ more than $$mathbb S$$ contains a closed topological copy of $$mathbb S$$. But $$F ( mathbb S)$$ also contains a topological copy of the square $$mathbb S times mathbb S$$ and so $$F ( mathbb S)$$ contains an unobtrusive discrete subspace. Is this situation typical?

Problem. Let $$G$$ to be a topological group containing a topological copy of the Sorgenfrey line. Is $$G$$ does it necessarily contain an unobtrusive discrete subspace?

## geometry – Are regular graphs always regular in a topological sense in a given dimension?

Consider a large cloud of points (sites) in arbitrary dimensions. Now, I introduce links between sites, so that any site is connected to exactly two other sites (and there are known self-connections or double connections). The resulting structure (finite graph, network or name of your choice) is now topologically equivalent to a simple 1D string (or several disconnected strings).

I wonder if this can also be generalized to higher dimensions? For example, if I link each cloud point site to exactly 3 (instead of 2) other sites, and for a while, ignore some of the subtleties such as the Handshaking lemma, is that also ok? ? always lead to a topologically regular structure (such as a two-dimensional honeycomb network, which has exactly the degree 3)?