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.

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|improve this question

2 Answers 2

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|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|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

 
discard

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