Define $A=\{(x,y):y=\sin(1/x), x\neq 0\}$ and $B=\{(0,y):-1\leq y \leq 1\}$.
How do I prove that $A\cup B$ is connected?
I can see this is not path connected but cannot prove why it is connected..
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Martin Sleziak
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user140374
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This seems to be very similar to this question: http://math.stackexchange.com/questions/426419/showing-the-topologists-sine-curve-is-connected-slight-variation (Although the statement of the other question is not entirely clear.) – Martin Sleziak Apr 19 '14 at 21:15
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Assume $X=A\cup B$ is the union of two open sets, $X=U\cup V$, with $U\cap V=\emptyset$. Wlog. $(0,0)\in U$. Then $U\cap A\ne\emptyset$ and $U\cap B\ne\emptyset$. As $A,B$ are (clearly, in fact pathwise) connected, we conclude $A\subseteq U$ and $B\subseteq U$, hence $V=\emptyset$.
Hagen von Eitzen
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+1, maybe the $U\cap A\neq \emptyset$, $U\cap B\neq \emptyset$ part could benefit from an additional argument. Unless it's "left as an exercise" of course. – Simon Markett Apr 19 '14 at 20:34
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@SimonMarkett I think the nonempty part is a bit straight from the definition of an "open" set. But the latter part $A\subseteq U$ and $B\subseteq U$ makes me think for a long time. – High GPA Nov 01 '23 at 10:19
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Let us deal first with the same problem for $A=\{(x,y); y=\sin\frac1x, x>0\}$. (Then you can get the set from the original problem simply as a union of two connected sets with a non-empty intersection.)
Notice that $A=\{(x,y); y=\sin\frac1x, x>0\}$ is connected - it is a continuous image of the connected space $(0,\infty)$.
If you notice that $A\cup B$ is precisely the closure of $A$, it suffices to show that closure of a connected set is again connected.
The proof of this lemma is not very difficult and it can be found in other posts on this site or elsewhere, for example:
- Prove that if $X$ and it's closure $\overline X$ are connected and if $X\subset Y \subset \overline X$, show that Y is also connected.
- Closure of a connected subset of $\mathbb{R}$ is connected?
- ProofWiki
- Ask a Topologist (Topology Q+A Board)
It is also useful to know that this is called Closed topologist's sine curve.
Martin Sleziak
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Hi Martin, I think that the topological sine curve $A\cup {(0,0)}$ is connected. But how to prove that $A$ is connected? – High GPA Nov 01 '23 at 10:34
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@HighGPA If we take a modified set corresponding to $x>0$, then we get a continuous image of a connected space. I have edited the answer a bit to clarify this. (I am not sure whether I actually noticed at the time that the question includes the negative values, too.) – Martin Sleziak Nov 01 '23 at 11:28