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.



Ends of topological spaces. Why independent of choice of ascending sequence of compact subsets?
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 $L\_{\*}$.


