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

Let $(X,\tau)$ be a topological space and $A$ a nonempty set. Let $X^A$ be the set of all mappings from $A$ to $X$. It is clear to me that the topology of pointwise convergence on $X^A$ is just the usual product topology on $X^A$ viewed as a Cartesian product.

However, let $\mathcal M$ be a nonempty proper subset of $X^A$. For example, attention is restricted on such mappings from $A$ to $X$ that satisfy some additional properties (say, continuity, when $A$ is topologized in some way), so that we no longer consider the set of all mappings from $A$ to $X$, just the set of some mappings from $A$ to $X$.

My question is: What is the proper interpretation of the notion of the “topology of pointwise convergence on $\mathcal M$?” Is it just the relative topology on $\mathcal M$ induced by the usual topology of pointwise convergence on $X^A$? Or is it usually defined some other way that differs from the relative topology?

share|cite|improve this question
up vote 1 down vote accepted

Yes, the topology of pointwise convergence on a subset $\mathcal{M} \subset X^A$ is the subspace topology induced on $\mathcal{M}$ by the product topology on $X^A$.

share|cite|improve this answer
I thought so, it does make sense, just wanted to make sure. Thank you very much for the quick answer! – triple_sec Dec 13 '13 at 20:40

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.