# How can I show that this shape is connected

Is the following shape connected? My intuition says yes...but I cannot prove it...any ideas please? $$\left\lbrace(x,y)\in\mathbb{R^2}:{x^2+y^2=1,0<x\le1}\right\rbrace\cup\left\lbrace{(x,y)\in\mathbb{R^2}: 0<x\le1,y=-1}\right\rbrace$$

• basic stuff from metric spaces... – dmtri Jan 5 '18 at 22:50
• I know about the example of the graph of $$sin(1/x)$$ but I cannot find something similar here... – dmtri Jan 5 '18 at 22:54
• this example looks similar to the disjoint sets $$(x,1/x)$$ and $$(x,0)$$ but in this case the union is disconnected... – dmtri Jan 5 '18 at 22:59

## 2 Answers

Tsemo appears to have answered the question in the comments beneath his post, so I'll just slightly generalize and summarize here to record a complete answer.

Suppose $A$ and $B$ are nonempty, connected subsets of a topological space $X$. Then $A\cup B$ is disconnected if and only if $\bar{B}\cap A = \bar{A}\cap B = \newcommand{\nullset}{\varnothing}\nullset$.

Proof: First suppose the latter condition holds. We'll prove this implies $A\cup B$ is disconnected. $X\setminus \overline{B}$ is open in $X$, so $(X\setminus\overline{B})\cap (A\cup B)$ is relatively open in $A\cup B$. But $$(X\setminus\overline{B})\cap (A\cup B) =(A\setminus\overline{B})\cup (B\setminus\overline{B}) =A\cup\nullset = A.$$ Hence $A$ is relatively open in $A\cup B$. Symmetrically $B$ is relatively open in $A\cup B$. Hence $A$ and $B$ disconnect $A\cup B$. (This is essentially the content of Tsemo's comment on his answer.)

Now suppose $A\cup B$ is disconnected, say by open sets $U$, $V$ of $X$ with $U\cap V \cap (A\cup B)=\nullset$, and $(U\cup V)\cap (A\cup B)=A\cup B$. Then $U\cap A$ and $V\cap A$ would disconnect $A$ if both were nonempty, but we assumed $A$ is connected, hence one of these is empty, and the other contains $A$. Suppose WLOG that $A\subset U$ and $A\cap V=\nullset$. Then similarly, since $B$ is connected, and we can't have $V\cap (A\cup B)=\nullset$, we must have $B\subset V$ and $B\cap U=\nullset$. Hence $B\subseteq U^C$, and $U^C$ is closed and disjoint from $A$. Thus $\overline{B}\cap A =\nullset$. Symmetrically, $\overline{A}\cap B = \nullset$.

The application to your question:

As for your case, the only point of intersection between the closures of the semicircle and the line segment is $(0,-1)$. This is not contained in either the semicircle or the line segment, hence the closure of the semicircle is disjoint from the line segment and vice versa. Thus the union is disconnected.

• I think there is no need for the A and B to be connected in the proposition...thanks again – dmtri Jan 5 '18 at 23:22
• @dmtri, it was necessary for the reverse direction. – jgon Jan 5 '18 at 23:24

draw a picture, it is not connected since it is the union of the half of a circle and a segment and they are disjoint.

• Can you please tell me, where to cut it? the segment is tangent to the circle... – dmtri Jan 5 '18 at 22:31
• Consider $\mathbb{R}^2$ minus the adherence of the segment and $V=\mathbb{R}^2$ -adherence of the half circle check that that $U$ contains the half circle and $V$ the segment. – Tsemo Aristide Jan 5 '18 at 22:36
• @uniquesolution, to whom goes the answer? – dmtri Jan 5 '18 at 22:43
• @uniquesolution But the two pieces are open in the subspace topology. That is, if $$X = \underbrace{\{(x,y) : x^2 + y^2 = 1 , x \in (0,1]\}}_{=:U_1} \cup \underbrace{(0,1]\times\{-1\}}_{=:U_2},$$ then both $U_1$ and $U_2$ are open in $X$. – Xander Henderson Jan 5 '18 at 22:44
• @uniquesolution, the question is obviously about the subspace topology, or it wouldn't make sense. – jgon Jan 5 '18 at 22:46