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I have a question about Example 2 of Section 25 (p.160) of Munkres's Topology.

Let $S$ be the following subset of the plane $\mathbb{R}^2$: $$S = \{ \ x \times \sin \tfrac{1}{x} \ \colon \ 0 < x \leq 1 \ \}.$$ Then $S$ is connected, being a continuous image of the connected subset $(0, 1]$ of $\mathbb{R}$, and the topologist's sine curve is the closure $\overline{S}$ in $\mathbb{R}^2$ of $S$ and is therefore also connected.

Let $V = \{0 \} \times [-1,1]$. Then $\overline{S} = V \cup S$.

Now Munkres states that the space obtained from $\overline{S}$ by deleting from $V$ all points $0 \times q$ such that $q \in \mathbb{Q}$, the set of rational numbers, is also connected.

How do we show that this is so? I know that the subspace $\mathbb{R} - \mathbb{Q}$ is not connected. I would prefer a rigorous argument, although an intuitive one would also be welcome.

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  • $\begingroup$ While it's great that you mention exactly where your questions come from, (or are inspired by) it would be best to do so in the body of your questions, while making your titles more descriptive of its contents. $\endgroup$ – user642796 May 6 '15 at 6:29
  • $\begingroup$ @ArthurFischer, you have a point, but the reason I prefer to do this is in order to be able to track my own questions later on and in order to help reduce duplication at the SE. You see, it would be more convenient for me to locate an answer to my questions if I get to read Munkres again in, say, two years' time and has unlearnt something I've learnt from the SE community this time around. $\endgroup$ – Saaqib Mahmood May 6 '15 at 6:35
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This is a special case of the following statement:

Suppose $A \subset X$ is connected. Then every subset $B \subset X$ such that $A \subset B \subset \bar{A}$, is also connected.

Proof: Suppose $U,V$ are open sets which form a separation of $B$. Since $B \subset \bar{A}$, we know that $U \cap A$ and $V \cap A$ are both nonempty. Therefore $U \cap A$, $V \cap A$ form a separation of $A$, contradicting the fact that $A$ is connected.

Now simply apply this statement when $A=S$ and $B = S \cup (\{0\} \times ([0,1] \backslash \mathbb{Q}))$.

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    $\begingroup$ Apparently you beat me by 8 seconds... $\endgroup$ – Nate Eldredge May 6 '15 at 5:31
  • $\begingroup$ Yeah sure no problem :) $\endgroup$ – Shalop May 6 '15 at 6:40
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Intuitively, even though $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q)$ is not connected (it is in fact totally disconnected), when you attach the topologist's sine curve it ''glues'' these points together, just in a different manner than adding in the rationals would glue them together.

More formally, let $U$ be a closed and open subset of $\overline{S} \setminus (\{0\} \times ([0,1] \cap \Bbb Q))$. Note that $U \cap S$ is closed and open in $S$, so it is either all of $S$ or disjoint from $S$. In the former case, the fact that $U$ is closed means it also contains every point of $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q$), and in the latter case the fact that it is open means it can contain no points of $\{0\} \times ([0,1] \cap \Bbb R \setminus \Bbb Q)$.

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  • $\begingroup$ I'm sorry, but I think you've been a bit too sloppy with the notation, especially the brackets and parentheses. $\endgroup$ – Saaqib Mahmood May 6 '15 at 6:30
  • $\begingroup$ @SaaqibMahmuud you are quite right, hopefully it is somewhat more readable now $\endgroup$ – Rolf Hoyer May 6 '15 at 6:33
  • $\begingroup$ yes, @Rolf Hoyer, it is more readable now. $\endgroup$ – Saaqib Mahmood May 7 '15 at 5:29

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