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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?

Many thanks for your help!

1 Answers 1

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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 complex 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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    @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.2011-04-22