# Why is a finite CW complex compact?

Hatcher explains on page 5 how a CW complex can be constructed inductively by attaching $n$-cells i.e. open $n$-dimensional disks.

On page 520 in the appendix he writes "A finite CW complex, ... , is compact since attaching a single cell preserves compactness."

Now my question: why is this obvious? An open disk is not compact, so how can I see that sticking two together is?

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You are attaching closed discs in a CW complex (In the notation of hatcher $D^n$ is the closed $n$-disc cf. page XII). Each closed disc is compact.

The CW compex is formed by taking a quotient of a compact space (the finite union of compacts is compact). Taking a quotient preserves compactness (Since the quotient space is the image of the original space under the projection map, and continuous maps preserve compactness).

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But on page 5 he writes "open $n$-disk" where he explains how to construct a CW complex in step (2). He is attaching open disks! – Rudy the Reindeer Apr 21 '11 at 10:25
@Matt- So usually the open disk' will correspond to what's called a cell,' but if you look carefully the way you actually construct the CW-complex is to glue the cell in along its boundary... i.e. look at a map $f: S^{n-1} \rightarrow X^{n-1}$ and use that to add an $n$-cell. – Dylan Wilson Apr 21 '11 at 10:44
He really attaches a closed disc to a pre-existing space. The boundary of this disc is identified with stuff that was already there, so the new points of this space is exactly the interior of the disc. In this sense he attaches an open disc. But mathematically he takes the quotient of the union of the existing space with a closed disc. – Thomas Rot Apr 21 '11 at 11:49
@Dylan: Thanks. I think now I understand. A cell is the interior of a closed disk. @Dylan: Did you mean to write "...is to glue a disk along its boundary..."? Because if a cell is open then it doesn't have a boundary. – Rudy the Reindeer Apr 22 '11 at 6:30