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Exercise: Let $A = \{(x, \sin (1/x)): 0<x\le 1\}$ and $B = \{(x,y)\in\mathbb R_{\le 0}\times\mathbb R | 0.5\le |y|\}$ be sets and $X = A\cup B$ the union. Show that $X$ is connected and path connected.

For connectedness: $A$ and $B$ are a cross-product of intervals so they are connected, but I don't how to prove the connectedness of the union.

Path connectedness: For $A$ it is clear. It is the path $\alpha (s)=(s,\sin (s^{-1})), s\in [a,b]$ for $a,b\in A$. But for the rest I don't know how to get the path. For explicit examples, it is clear that $X$ is path connected but I find no general solution.

Generally It is enough to prove the connectedness with path connectedness.

PVAL-inactive
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serge
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  • see http://math.stackexchange.com/questions/35054/topologists-sine-curve and http://math.stackexchange.com/questions/317125/topologists-sine-curve-is-connected – Ivica Smolić Jan 18 '16 at 18:31
  • Okay, with following Lemma:If $A$ and $B$ are connected, and $A\cap B\neq\emptyset$, then $A\cup B$ is also connected. The first part is done, because $A,B$ are disjoint and connected hence there union is connected. But for the seccond part, I don't know how to apply the examples from the link. – serge Jan 18 '16 at 18:44

1 Answers1

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I am assuming that $B= (-\infty,0]\times \{y\in R: |y|\geq 1/2\}.$ Then $A\cup B$ is NOT path-connected.

Proof : Assume that $f:[0,1]\to A\cup B$ is continuous with $f(0)=(1,\sin 1)$ and $f(1)\in B.$

Let $r=\min \{t\in (0,1] :f(r)\in B\},$ which exists because $f^{-1}B$ is closed in $[0,1]$, and $f(0)\not \in B.$

Let $D =\{f(t): t\in [0,r)\}.$

$\bullet$ We have $ D=A$ because

  1. $f(r)\in B$ and the continuity of $f$ require that $f(r)\in Cl_{A\cup B} D$, with $D\subset A,$ so $\inf \{x\in [0,r) :(x,\sin 1/x ) \in D \}=0,$ and

  2. $ D$ is a connected space as it is a continuous image of $[0,r),$ so if $(z,\sin 1/z)\in A\backslash D$ then $\{(x,y)\in D :x<z\}$ and $\{(x,y)\in D :x>z\}$ are disjoint non-empty subsets of the space $ D,$ and their union is $D,$ contradicting connectedness of $D$.

$\bullet$ Now let $f(r)=(0,s).$ We have $1/2\leq |s|\leq 1$ because the continuity of $f$ implies $(0,s)\in (Cl_{A\cup B} D)\cap B=(Cl_{A\cup B} A)\cap B=\{(0,y): 1/2\leq |y|\leq 1\}.$

$\bullet$ Consider the function $g:[0.1]\to A\cup B$ where $g(t)=f(t r)$ for $t \in [0,1]$. Now $g$ is a path from $(1,\sin 1)$ to $(0,s)$ with $\{g(t): t\in [0,1)\}=D=A.$

$\bullet$ And consider the nbhd $V=(A\cup B)\cap \{(x,y): |x|<1/4 \land |y-s|<1/4\}.$ There exists $d>0$ such that $t\in (1-d,1]\implies g(t)\in V.$

$\bullet$ Now the space $A\cap V$ has infinitely many connected components, and each connected component $C$ of $A\cap V$ satisfies $B\cap (Cl_{A\cup B} C )=\phi.$ But since $E=\{g(t):t\in (1-d,1)\}$ is connected, there must be a component $ C$ of $V\cap A$ such that $E\subset C.$ This implies that $g(1)=f(r)=(0,s)\not \in Cl_{A\cup B}E,$ that is, $g(1)\not\in Cl_{A\cup B}\{g(t):1-d<t<1\},$ contradicting the continuity of $g.$ QED.

DanielWainfleet
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  • Anyone welcome to edit for better layout . I'm too tired. – DanielWainfleet Jan 20 '16 at 05:31
  • $A$ is connected because it is homeomorphic to $(0,1)$ and $A\cup B$ is connected because $A$ is a connected dense open subspace of $A\cup B$. – DanielWainfleet Jan 20 '16 at 06:47
  • What I meant to say in my previous comment is that $ A$ is a connected dense open subspace of $ F=Cl_{A\cup B}A$, so$ F$ is connected, from which it is easily seen that $A\cup B$ is connected. – DanielWainfleet Jan 20 '16 at 15:30