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I have a quick question about the difference between the two concepts in the title. The question is basically ex.6 (b) in Hatcher's book titled "Algebraic Topology". Let $X$ be the subspace of $R^2$ consisting of the horizontal segment $[0,1] \times \{0\}$ together with the vertical segments $\{r\} \times [0,1-r]$ for $r$ a rational number in $[0,1]$. Now let $Y$ be the space that is the union of an infinite number of copies of $X$ arranged in a zig zag formation. See below -

The space $Y$

Now my question is why can't one deformation retract $Y$ to a point in the darkened zig zag line? Surely the darkened zig zag line is homeomorphic to $\mathbb{R}$, which is deformation retractable to a point, and each of the vertical lines of each copy of $X$ deformation retracts to its segment of the zig zag line! I must be missing something here as one has to prove that $Y$ does not deformation retract to any point!

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marked as duplicate by Najib Idrissi, SchrodingersCat, Michael Medvinsky, Alex M., Leucippus Dec 19 '15 at 15:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 14 down vote accepted

In exercise 5, that's the one before this one, you showed that "if a space $X$ deformation retracts to a point $x \in X$, then for each neighbourhood $U$ of $x$ in $X$ there exists a neighborhood $V \subset U$ of $x$ such that the inclusion map $V \hookrightarrow U$ is nullhomotopic."

Pick a point $z$ in $Z$ (the zig zag line). Then you can find a neighbourhood $N$ of $z$ that is disconnected and such that every neighbourhood $U$ with $z \in U \subset N$ is also disconnected. Then you apply 5.

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Doesn't this answer also give a much quicker answer to this question? – t.b. May 28 '12 at 14:12
@t.b. It looks like it. But I might be missing something. – Rudy the Reindeer May 28 '12 at 14:19
I see. The other question is about contractibility, not about (strongly) deformation retracting to a point. Your argument only shows that the latter can't be while the former could still be the case. – t.b. May 28 '12 at 14:25
Thanks Matt N. for submitting an answer. – Sean May 28 '12 at 15:26
The mathematically correct answer to Sean's question given by Rudy is clear, but what is the intuition? Sean's argument, although informal, seems correct. Where is Sean's intuitive argument incorrect? – user54301 Dec 15 '15 at 22:39

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