# Tag Info

2

How if we have a neighborhood $U$ of $1$, with $\overline{U^{-1}}$ compact? It seems the following. The answer is positive, because in this case $G$ is a (locally compact) topological group. So it has a base at the unit consisting of symmetric $(U=U^{-1})$ neighborhoods. Now we prove the first claim. A family $\{xU:x\in \overline{U^{-1}}\}$ is an ...

5

The proof of contractibility in the strong(=weak=compact-open) topology is very easy: Identify $H$ with $L^2([0,1])$. The path $\{u_t\}$ that connects any unitary $u$ to the identity element is then given by: $$u_t:L^2([0,1])= L^2([0,t])\oplus L^2([t,1])\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$ $$\qquad\qquad\qquad \xrightarrow{\varphi_t u ... 10 If A is open and B is any set, then$$AB = \bigcup \{Ab : b \in B\} is open (hence Borel). If $T$ is Borel, then $AB \cap T$ is an open subset of $T$, hence Borel. You may already know this, but if you weaken "$A$ is open" to "$A$ is $G_\delta$", then $AB$ does not have to be Borel. This result can be found in "On the sum of two Borel sets" by Erdos ...

9

Here's another counterexample for Problem 1 not using the fancy (non-explicit) construction of Shelah. Fix an odd prime $p$. Let $A=\mathbf{F}_p((t))$ be the ring of Laurent series over the field $\mathbf{F}_p$ on $p$ elements; we only view it as an abstract group, and let $A_0$ be the set of series $\sum a_nt^n$ in $A$ with $a_0=0$. Let $B$ be the ...

3

This is possible. Suppose that $\mu(W)<\infty$ and that $\pi(W)$ is measurable. Then $\mu(W)$ is the supremum of all compact sets $K$ contained in $W$, as $\mu$ is a Radon measure. The same holds for $\pi(W)$. For a compact set $K$ the measure $\mu(K)$ is the infimum of all integrals $\int_Gf(x)d\mu(x)$ where $f\ge 0$ is continuous of compact support. ...

13

Counterexample for Problem 1: According to this answer Saharon Shelah constructed a "Jónsson group" of order $\aleph_1,$ i.e., an uncountable group $G$ in which every proper subgroup is countable. Choose an element $g\in G\setminus\{e\}.$ Let $H$ be a maximal subgroup of $G$ not containing $g.$ Since $H$ is a proper subgroup of $G,$ $H$ is countable. By the ...

0

According to this paper on open problems in topological algebra, this is an open problem (it's listed as Problem 9 in particular). Problem 9 (Protasov). Can every topological group be algebraically generated by a nowhere dense subset? Let us mention that each countable topological group is algebraically generated by some closed discrete subset ...

Top 50 recent answers are included