# Ends of topological spaces. Why independent of choice of ascending sequence of compact subsets?

Quoting from http://en.wikipedia.org/wiki/End_(topology):

"Let X be a topological space, and suppose that

K1 ⊂ K2 ⊂ K3 ⊂ · · ·

is an ascending sequence of compact subsets of X whose interiors cover X. Then X has one end for every sequence

U1 ⊃ U2 ⊃ U3 ⊃ · · ·

where each Un is a connected component of X \ Kn. The number of ends does not depend of the specific sequence {Ki} of compact sets; in fact, there is a natural bijection between the sets of ends associated with any two such sequences."

How does one prove that the number of ends does not depend on the specific sequence of {Ki} of compact sets?

An expository proof for a relatively neophyte math student (no category theory...) would be much appreciated.

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Perhaps you should reconsider your tags. – Martin Brandenburg Mar 5 '11 at 13:44
Consider $invlim_{K \subset X} \pi_0 (X-K)$, where $K$ ranges through all compact sets, ordered by inclusion. An exhaustion $K_1 \subset K_2 ...$ as in wikipedias definition is a final sub-poset of the poset of all compact subsets, thus $invlim_{K \subset X} \pi_0 (X-K) \to invlim_n \pi_0 (X-K_n)$ is a bijection. An element in the target is nothing but an end of $X$. Given two compact exhaustions, you combine the two bijections obtained in this way. Admittedly, this is a bit categorical, but wringing out the categories is an exercise with inverse limits. – Johannes Ebert Mar 5 '11 at 13:48

The main point is this. Let $(L_k)_{k=0}^\infty$ be another increasing sequence of compact sets whose interiors cover $X$. Each $X_n$ is compact and contained in the union of the sets $\text{int}(L_k)$, so it is contained in some finite union of these open sets. As the sets $\text{int}(L_k)$ are nested, it follows that $K_n\subseteq\text{int}(L_{k_n})\subseteq L_{k_n}$ for some index $k_n$. Moreover, we can assume that $k_n < k_{n+1}$. Similarly, there is an increasing sequence of indices $n_k$ such that $L_k\subseteq\text{int}(K_{n_k})\subseteq K_{n_k}$ for all $k$. Now each component of $X\setminus L_{k_n}$ is contained in a unique component of $X\setminus K_n$, and each component of $X\setminus K_{n_k}$ is contained in a unique component of $X\setminus L_k$. This gives maps $\pi_0(X\setminus L_{k_n})\to\pi_0(X\setminus K_n)$ and $\pi_0(X\setminus K_{n_k})\to\pi_0(X\setminus L_k)$, which can be assembled to give a bijection between the set of ends defined using $K_{*}$ and the set of ends defined using $L_{*}$.

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Thanks, but, and sorry for being so stupid, it still isn't quite clear for me. I can't see why we can assume that $k_n< k_{n+1}$. Why cannot, or must not, $k_n≤ k_{n+1}$? And therefore, each component of $X\setminus L_{k_n}$ is not contained in a unique component of $X\setminus K_{n}$, and vice versa? – Simon Mar 6 '11 at 21:57
The main property that you need for $k_{n+1}$ is that $K_{n+1}\subseteq\text{int}(L_{k_{n+1}})$. If there exists a value of $k_{n+1}$ with this property, then any larger value will also have the property. Thus, if the first $k_{n+1}$ that we thought of was not strictly larger than $k_n$, we could just choose to make it bigger. – Neil Strickland Mar 6 '11 at 22:10
Thank you, I realize now that I had assumed $k_{n}$ to be minimal for each $n$ such that $K_{n}\subseteq\text{int}(L_{k_{n}})$. Being new to typology, most are the details a latent risk for misconception, I suppose. – Simon Mar 7 '11 at 23:18

Neil has already given adequate reply; this answer is partly for Simon, and partly for those who do like category theory, and realize that its purpose is to make life simpler, not more complicated!

First, IMHO that's not a very good definition in the wikipedia article. A better definition is given in Spivak's A Comprehensive Introduction to Differential Geometry, Volume I, page 30: an end of a non-compact topological space $X$ is a function $e$ which assigns to each compact subset $K \subset X$ a nonempty component $e(K)$ of the complement $X - K$, in such a way that $K \subset K'$ implies $e(K') \subset e(K)$. This way of putting it circumvents having to choose a covering by interiors of compact sets at the outset, and then requiring a lemma that shows independence of choice.

In categorical language, the set of ends of $X$ is the inverse limit of sets

$$\lim_{K \subset X} \pi_0(X - K)$$

where $K$ ranges over compact subsets.

Anyway, in answer to the question, the point is that any sequence of compact subsets whose interiors cover $X$ is cofinal in the directed set of all compact subsets. (A partially ordered set is directed if it is nonempty and if any two elements have an upper bound. A subset is cofinal if any element in the partial order is bounded above by an element in the subset.)

The point then is that the limit over a directed set is isomorphic to the limit over a cofinal subset (with partial order inherited from the order of the directed set): in the present case, the sequence $K_j$ is cofinal, and the map given by restriction

$$\lim_{K} \pi_0(X - K) \to \lim_j \pi_0(X - K_j)$$

is a bijection. The inverse function takes a sequence of components $C_j$, and assigns to it the function whose value at $K$ is the unique component of $X - K$ which contains $C_j$, where $K_j$ is any compact subset containing $K$. This doesn't depend on $j$, and it is routine to show this does give the inverse function, according to what Neil has already explained.

But it is really just a special case of a much more general argument about cofinal functors; see Categories for the Working Mathematician, page 217, for a rather more general statement.

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