Open Sets in the Product Toplogy

**mpetnuch** · December 6th 2007, 07:47 PM

I have a problem that is two fold. First, I am not comfortable with the notation that my textbook uses. I am familiar with another notation of product space (from say Munkres which I read on my own first). Secondly, I am not really comfortable wit the product toplogy. I can get away most of the time, i.e., when the products are finite and hence equal to the box topology but when I must use the product topology I get a headache :-) So I am hoping that maybe somebody here could help me.

First off here, is the begining of the problem I am trying to solve:

Let $\mathcal{K}$ be a set of continuous functions from $(X, \tau)$ into $[0,1]$ .

Define $K : X \to [0,1]^\mathcal{K}$ by $K(x)(f) = f(x)$ .

I then translated it into what I think is right in notation that I am familiar with. Here it is:

Consider $\{ f_\alpha \}_{\alpha \in J}$ , an indexed set of continuous functions $f_\alpha : X \to [0,1]$ and define $f : X \to [0,1]^J$ by $f(x) = (f_\alpha(x))_{\alpha \in J}$ .

Assuming that is right, my next problem is to show that $K$ or in my notation $f$ is continuous. But what are the open sets in $[0,1]^J$ . I think they are $\pi^{-1}_\alpha(U)$ where $U$ is open in $[0,1]$ ? Are these open sets functions? I understand that the "points" in $[0,1]^J$ are functions (i.e. the $f_\alpha$ 's), but what are the open sets. Are they an uncountable number of continuous functions. Let me stop here before I start sounding really dumb.

Thanks for the help.

**Rebesques** · December 7th 2007, 06:45 AM

If I understand what you are saying, you just pick open sets $A_j\subset [0,1]$ and show that the set $f^{-1}(\times_j A_j)$ is open in $X.$

To this end, notice that $f^{-1} (\times_j A_j)=\cup_j f_j^{-1} (A_j)$ and that the $f_j$ are continuous.

**Opalg** · December 7th 2007, 07:52 AM

The way to get your head round the product topology is to remember that a neighbourhood of a point can only control finitely many of the coordinates of that point.

But before you can consider neighbourhoods, you have to be clear about what a point in the product space looks like. A point z in $[0,1]^{\mathcal{K}}$ is a collection of coordinates $z=(z_f)_{f\in\mathcal{K}}$ indexed by the functions in $\mathcal{K}$ (where $z_f\in[0,1]$ for each coordinate z_f).

A typical neighbourhood of z in the product topology is the set N(z) consisting of all elements $w=(w_f)\in [0,1]^{\mathcal{K}}$ which differ from z by less than ε at a finite number of coordinates, say the coordinates $f_1,\:f_2,\,\ldots,f_n$ . Thus $N(z) = \{w\in [0,1]^{\mathcal{K}} : |w_{f_j} - z_{f_j}|< \varepsilon\;(1\leqslant j\leqslant n)\}$ .

Now we can start to look at the map $K:X\to [0,1]^{\mathcal{K}}$ , which is defined by $(K(x))_f = f(x)$ . The condition for K(y) to lie in the neighbourhood N(K(x)) is $|(K(y))_{f_j} - (K(x))_{f_j}|<\varepsilon$ for 1≤j≤n, or in other words $|f_j(y)-f_j(x)|<\varepsilon$ . To prove that K is continuous, we have to show that this condition will hold whenever y is close enough to x in the space X. But since the condition only involves finitely many functions $f_j\in\mathcal{K}$ , and these functions are all continuous, it should be easy enough to see that this can be done.

Thread: Open Sets in the Product Toplogy

LinkBack

Thread Tools

Display

Open Sets in the Product Toplogy

Similar Math Help Forum Discussions

Open sets and sets of interior points

Metric spaces, open sets, and closed sets

Prove: The intersection of a finite collection of open sets is open in a metric space

An open set is disconnected iff it is the union of two non-empty disjoint open sets

Is the product of two open sets open?

Search Tags