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I think this is done inductively on the skeletons but I can't work out the details.

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    They're not just locally path-connected, they're locally contractible. There's a key theorem about CW-complexes, that the inclusion of any of any subcomplex into the entire CW-complex is a cofibration. Look at that proof and the neighbourhoods constructed in that proof. That should give you the idea for how to prove what you want to prove.2011-06-19
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    Please give full details in the question, not just the title. Also, what is a CW complex?2011-06-19
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    @Asaf What additional details do you want, exactly? And all definitions are easily googlable.2011-06-19
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    @Grigory: It's not that I complain about lack of definitions. I complain about bad formatting, while at it I was asking what is a CW complex.2011-06-19
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    @Asaf: First google hit: [CW complex](http://en.wikipedia.org/wiki/CW_complex): A space obtained by gluing disks together. The topologist's preferred notion of a polyhedron. All reasonable (geometric) spaces are CW complexes. C: closure finite, W: weak topology. Inventor: J.H.C. Whitehead.2011-06-19
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    Have you looked at Hatcher's book?2011-06-19
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    Dear @RyanBudney I saw this problem in Lee's book where he haven't introduce the conception of contractible. Can you tell me how to prove it directly? Thank you!2014-07-17
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    I proved this very carefully here https://math.stackexchange.com/questions/1448789/understanding-construction-of-open-nbds-in-cw-complexes/1462419#14624192017-06-18

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You can use the following two general topology facts:

  1. A disjoint union of locally path-connected spaces is locally path-connected.
  2. A quotient of a locally path-connected space is locally path-connected (See Lemma 2 in this post).

Let $X_n$ be the n-skeleton of a CW-complex $X$. $X_0$ is obviously locally path-connected. Inductively, $X_n$ is the quotient of the disjoint union of $X_{n-1}$ and n-cells so $X_n$ is locally path-connected. Now $X=\bigcup_{n}X_n$ has the weak topology with respect to the subspaces $X_n$. Equivalently, $X$ has the quotient topology with respect to the map $\coprod_{n}X_n\to X$ defined in terms of the inclusions. Since $\coprod_{n}X_n$ is locally path-connected, so is $X$.