Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A metrizable group is a metric space $(G,d)$ with a binary operation $\cdot$ such $(G,(\cdot))$ is a group and maps $(\cdot):G\times G\to G$ and $f:G\to G$ given by $(\cdot)(x,y)=xy$ and $f(x)=x^{-1}$ are continuous respect $d$.

Why requires definition that $f$ be continuous? Is it possible to have continuity of $(\cdot)$ without continuity of $f$?

share|improve this question
1  
I don't understand your last sentence. Of course, $y^{-1}$ exists; the question is whether the inversion map is continuous. –  Ted May 31 '12 at 3:21
    
But.. inverses are related with the binary operation, I don't know why need an aditional condition. –  Gastón Burrull May 31 '12 at 3:24
2  
(i) Actually, formally a group is defined by three operations: a binary operation (the product), a unary operation (inversion), and a nullary operation (the identity element). (ii) Are you asking whether the continuity of inversion follows from continuity of the product? –  Arturo Magidin May 31 '12 at 3:26
1  
Yes, they are "related", but are they "related" enough that the continuity of the binary operation implies the continuity of the inversion map? I think the answer is no although I don't have a counterexample at hand. –  Ted May 31 '12 at 3:26
1  
@GastónBurrull: First, no, they don't "induce new operations". They are part of the definition of a group when you describe a group as an algebra. Second: the point is that if you really want to define a topological group, you need to consider all operations as functions; the inverse function cannot be expressed that way in terms of the product function alone (you need existential quantifiers to describe it). The nullary function is continuous because it is constant, so you don't need to put conditions on it. So your question is really "does continuity of $\cdot$ imply that of $f$?" –  Arturo Magidin May 31 '12 at 3:41
show 4 more comments

2 Answers

up vote 6 down vote accepted

I made a mistake here before. Sorry!

Continuity of multiplication is not sufficient for general topological groups.

Consider the standard additive group of reals with Sorgenfrey topology. Inverse is clearly not continuous, since $-[a,b)=(-b,-a]$.

Multiplication (or, rather, addition) is continuous, since, intuitively, the open sets in the product topology are the sets which do not have "upper" and "right" edge (but may possibly have "left" and "lower" edge, or some parts of those). More precisely, for any $x+y=c\in [a,b)$, then let $0<\varepsilon<(b-c)/2$. Then for any $(x',y')\in [x,x+\varepsilon)\times [y,y+\varepsilon)$, $a\leq x+y\leq x'+y' <x+y+\varepsilon<b$, so $(+)^{-1}[a,b)$ is open.

EDIT: I found an example in Topological Groups and Related Structures by Archangel'skii & Tkachenko, example 3.5.6 on pages 175-176. It is the group of homeomorphisms of the space $\lbrace n,0,1/n\mid n\in \mathbf N\rbrace$ with the natural topology, the compact-open topology.

It is metrizable, but not metrizable by a left- or right- invariant metric, and the inverse is not continuous.

share|improve this answer
    
I need more time to full understand your answer. I'll acept when I'll totally sure. Thanks! –  Gastón Burrull May 31 '12 at 4:52
1  
I think things are OK for completely metrizable spaces. Forget where I saw it. –  André Nicolas May 31 '12 at 5:12
5  
To follow up on André's comment, continuity of inversion follows from continuity of multiplication in quite some generality by a Baire category argument, e.g. for locally compact Hausdorff or completely metrizable spaces with a group structure such that multiplication is (separately) continuous: Pfister, Continuity of the inverse, Proc. Amer. Math. Soc. 95 (1985), 312-314 contains the best results in this direction I'm aware of. –  t.b. May 31 '12 at 8:38
    
@t.b. Thanks a lot!!. Must you mean that be metrizable is not sufficient condition?. –  Gastón Burrull May 31 '12 at 22:28
2  
@Gastón: No, metrizability is not a sufficient condition for continuity of the multiplication to imply continuity of the inversion. tomasz made an edit to the answer giving a reference to a counterexample. –  t.b. Jun 1 '12 at 3:41
show 1 more comment

Take $G=\mathbb R$ with addition and put on it the topology which has the set $\{(-\infty,a):a\in\mathbb R\}$ as a basis. Then addition is continuous but inversion is not.

You want it metrizable, though...

share|improve this answer
    
Probably your example is the easier when inversion continuity not follow from other in topological spaces. Sometimes I think that metrizable spaces are not enough studied like other topological spaces. –  Gastón Burrull May 31 '12 at 22:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.