The topologist's sine curve is the closure $\overline{S}$ of the subset $S$ of $\mathbb{R}^2$, where $$S\ = \ \{ \ x \times \sin \frac{1}{x} \ \colon \ 0 < x \leq 1 \ \}.$$ So $$\overline{S} = S \cup V,$$ where $$V = \{0 \} \times [-1,1].$$

Now let $A$ the subset of $V$ obtained by removing all points with rational second co-ordinates; that is let $$A = \{0\} \times \left( [-1,1] - \mathbb{Q} \right). $$

Then $A$ is not connected as a subspace of $V$. Am I right?

Is $A$ connected as a subsapce of $\overline{S}$?

  • 1
    $\begingroup$ Regarding question 2: It looks like $A$ is not a subspace of $S$ at all; am I misunderstanding something? $\endgroup$ – Eric Stucky May 6 '15 at 7:03
  • $\begingroup$ @EricStucky, you're absolutely right. I'll edit my post forthwith. $\endgroup$ – Saaqib Mahmood May 6 '15 at 7:48
  • $\begingroup$ I've made a correction to my original post. $\endgroup$ – Saaqib Mahmood May 6 '15 at 12:19

Suppose $A$ is connected, and consider the inclusion mappings $A \to V \to \Bbb{R^2}$. These maps are continuous, so you conclude that $A$ is connected also as a subspace of $\Bbb{R}^2$. Contradiction.

  • $\begingroup$ $A$ is not connected in $V$. Can't we use the same argument with $V$ replaced by $\overline{S}$ to conclude that $A$ is not connected in $\overline{S}$? $\endgroup$ – Saaqib Mahmood May 6 '15 at 7:51
  • $\begingroup$ Yes. Crostul's answer is actually exactly the same as that, but with a fancy wrapper around it. $\endgroup$ – Eric Stucky May 6 '15 at 7:53
  • $\begingroup$ Oh, yes, this argument applies on every set containing $A$. $\endgroup$ – Crostul May 6 '15 at 7:55

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