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.

Given an inverse system $\mathcal G=\{X_i\}$ of topological spaces over some directed set $I$.

If $X=\prod\limits_{i\in I}X_i$, the inverse limit $X^*=\varprojlim X_i$ of $\mathcal G$ is a subspace of $X$

Could someone explain this to me in a very basic way (I have read many references but could not get it). How $X^*$ is a subspace of $X$.

share|improve this question

3 Answers 3

up vote 3 down vote accepted

The product $\prod_{i\in\omega}X_{i}$ is the set of all sequences $(x_{0},x_{1},...)$ where $x_{i}\in X_{i}$ for each $i$ .

Now you are given a family of continuous “bonding” maps $f_{i}:X_{i+1}\to X_{i}$ .

The inverse limit is a special subset of $\prod_{i\in\omega}X_{i}$ , consisting of those sequences $(x_{0},x_{1},...)$ satisfying $f_{i}(x_{i+1})=x_{i}$ for each $i$ . I like to think of the elements of the inverse limit as “threads” because the terms are linked together by the functions $f_{i}$ . If you picture this from $X_{0}$ going up, you get a tree-like structure.

Now give this subset the subspace topology, where $\prod_{i\in\omega}X_{i}$ has the product topology. That is, $U$ is open in $X^*$ iff $U=V\cap X^*$ for some open $V$ in $\prod_{i\in\omega}X_{i}$.

share|improve this answer
    
Excellent, I get the idea. –  Layman Mar 14 at 0:54
2  
Notice that this is the special case $I=(\mathbb{N},\leq)$. –  Martin Brandenburg Mar 14 at 1:08

You just construct $\varprojlim_i X_i$ as a subspace of the product $\prod_i X_i$, namely consisting of those tuples which are compatible with respect to the transition maps. This even works for every index category, not just for directed partial orders.

share|improve this answer
    
I am sorry I don't understand it. could you please construct it as an example or for finite index. –  Layman Mar 13 at 22:20
    
Why should I do this? Nothing becomes easier in this special case. Please ask specific (!) questions - so that we can help efficiently. –  Martin Brandenburg Mar 13 at 22:34
    
My question is very clear. How to prove $X^*$ is a subspace of $X$. I cannot digest the way you have written. more simpler if you can! –  Layman Mar 13 at 22:54
    
How do you define and construct $X^*$? –  Martin Brandenburg Mar 13 at 23:30
1  
1. Recall the definitions (topological spaces, subsets, subspaces)! 2. Given a subset of the underlying set of a topological space, one usually endows it with the subspace topology. This way, it becomes a subspace. There is nothing to prove here! If you have troubles with this, you shouldn't deal with inverse limits, but with very very basic constructions of metric and topological spaces first. Otherwise you will have many other troubles, soon. –  Martin Brandenburg Mar 14 at 1:09

As you tag (category-theory), I will assume you have basic knowledge about it.

The forgetful functor $\mathsf{Top} \to \mathsf{Set}$ admits a left adjoint (namely the discrete topology functor $\mathsf{Set} \to \mathsf{Top}$), so it commutes with (small) limits. Then the underlying set of the limit in $\mathsf{Top}$ is the limit in $\mathsf{Set}$ of the underlying space.

Now, either you know that in $\mathsf{Set}$, every limit $\lim_J F$ is a subset of $\prod_{j \in \mathrm{Ob}\, J} F(j)$, either you show it (once you know that you search such a subset, it is not that hard to guess it ; then show it is indeed the limit).


Edit. Ok, you did not tag (category-theory), so my apologies if it is obscure to you. I leave it nevertheless for those whose land on this page and would be interesting in such an answer.

share|improve this answer

Your Answer

 
discard

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.