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I am working on Exercise 4.1.11 in Hatcher's Algebraic Topology text, which is as follows:

Show that a CW complex $X$ is contractible if it is the union of an increasing sequence of subcomplexes $X_1 \subset X_2 \subset \cdots$ such that each inclusion $X_i \hookrightarrow X_{i+1}$ is nullhomotopic.

By applying Whitehead's Theorem, if we show that $\pi_n(X,x_0) = 0$ for all $n$ and any point $x_0 \in X$ and that $X$ is connected, it will follow that $X$ is contractible.

It is already shown here that $\pi_n(X,x_0) = 0$ for all $n$ and any point $x_0 \in X$. What's not shown there, however, is that $X$ is connected.

How can we show that $X$ is connected?

Thanks!

michiganbiker898
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1 Answers1

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If you know that $\pi_0(X,x_0)=0$, then the result is trivial. Explicitly, you can consider two points $x,y$. There is an $i$ such that $x,y\in X_i$. Since the inclusion $X_i\to X_{i+1}$ is nullhomotopic, there is an arc connecting these two points in $X_{i+1}$.

Sergey Guminov
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    Thanks! So, triviality of $\pi_0(X,x_0)$ for any point $x_0 \in X$ corresponds to path-connectedness (and hence connectedness) of $X$, right? – michiganbiker898 Jan 07 '22 at 21:31
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    Aah, I see. What I said in the comment above is correct: the discussion at https://math.stackexchange.com/questions/1596822/zeroth-homotopy-group-what-exactly-is-it is helpful for this. – michiganbiker898 Jan 07 '22 at 21:34