Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As part of a casual self-study of topology, I have started fooling around with topological groups. I noticed that the article on Wikipedia mentioned that "weakening the continuity conditions" gives the definition for a semitopological group as one in which the functions $l_g,r_g:G\rightarrow G$ defined by $l_g(h)=gh$ and $r_g(h)=hg$ are continuous for each group element $g$.

Since this is supposedly a weaker condition, every topological group should be semitopological, but I don't understand how one would go about proving this. I feel like I should be able to use some sort of relationship between continuous functions from $G\times G$ to $G$ and continuous functions from $G$ to $G$, but I haven't been able to find anything of that nature.

So how can it be proven that topological groups are semitopological?

share|cite|improve this question
up vote 3 down vote accepted

The restriction of a continuous function to a subspace is continuous. More generally, the composition of continuous functions is continuous (and restricting to a subspace is the special case that one of the continuous functions is the inclusion of a subspace).

share|cite|improve this answer

Basically, if the multiplication is continuous as a function of its two variables (as in a topological group) then the multiplication is continuous when you fix one of the variables and consider the resulting function on the other variable (as in a semitopological group). Informally, it is easier to be continuous with respect to just varying one of the two variables, hence every topological group is a semitopological group. For a proof, just notice that this is a case of restricting a function.

share|cite|improve this answer

Your Answer


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.