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

I'm trying to understand the topology of pointwise convergence, we're defined it on the set $\cal{F}(X)$ of real functions on set $X$ to be the sub basis with topology $$\{f \in \cal{F}(X) : a < f(x) <b\} $$ where $x \in X$ and $a,b \in \mathbb{R}$.

Then it says: 'A set from the sub-basis consists of all functions that pass through one vertical interval'. But at what $x$ value is this interval? And which out of $x,a,b$ are 'changing' to create the sub basis?

share|cite|improve this question
up vote 2 down vote accepted

Any $x\in X$ and any pair $a<b$ of real numbers gives rise to such a set $$S(x, a, b) = \{f:\, a<f(x)<b\}$$ The set of all such $S(x, a, b)$ with $x\in X, a < b$ real, is the subbasis of your topology.

share|cite|improve this answer

The topology of pointwise convergence is the same as the product topology on $\mathbb R^X$.

A (sub)basis for the product topology is the set $\{ \prod_{j \in J \subset X} O_j \times \prod_{i \in X \setminus J} \mathbb R \mid O_j \subset \mathbb R \text{ open }, J \text{ finite } \}$.

share|cite|improve this answer
I thought I'd add that even though it doesn't directly answer your question. – Rudy the Reindeer May 29 '12 at 17:18

Your Answer


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