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The topologist sine curve is given by taking the closure of $$S = \{ x \times \sin (1/x) \, | \, 0 < x \le 1 \}$$ in $\mathbb{R}^2$. So it is $\bar{S} = S \cup ( \{ 0 \} \times [-1,1] )$.

I was reading the following proof (James Munkres' pg 157) showing that it is not path connected, which begins with,

Suppose there is a path $f:[a,c] \rightarrow \bar{S}$ beginning at the origin and ending at a point of $S$. The set of those $t$ for which $f(t) \in \{0 \} \times [-1,1]$ is closed, so it has a largest element $b$...

This is the part where I do not understand. Why must $f^{-1} (\{0 \} \times [-1,1] )$ be a closed set? All we know is that $f$ is continuous. Sorry if I am missing something trivial, thanks!

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$\{0\} \times [-1, 1]$ is closed in $\overline S$ (since $\{ 0\} \times [-1,1]$ is closed in $\mathbb R^2$ and you take the subspace topology of $\overline S$) and $f$ is continuous, so $f^{-1}(\{0\}\times [-1,1])$ is closed (by definition of continuity of $f$)

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  • $\begingroup$ Do you know the justification for the next statement, "... so it has a largest element $b$."? $\endgroup$ – user486983 Jan 25 '18 at 1:57
  • $\begingroup$ That's because closed and bounded set in $[-1,1]$ must have a largest elements (that's a consequence of Bolzano- Weierstrass theorem) @Ella $\endgroup$ – user99914 Jan 25 '18 at 3:06
  • $\begingroup$ Did you mean compact sets in $\{0\}\times [-1,1]$? $\endgroup$ – user486983 Jan 26 '18 at 8:06

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