# Proving that $\{(x,y):x^2-2x+y^2=0\}\cup \{(x,0):x\in [2,3]\}$ is connected

Let $E=\{(x,y):x^2-2x+y^2=0\}\cup \{(x,0):x\in [2,3]\}$. It appears to me from the sketch that this set is connected. However, I have no idea how to prove this, since the set is not convex, and we probably cannot use the line connecting any two points of the set argument to show the set is connected. I have almost no experience proving things like that, would appreciate some help.

Note that a union of two connected sets is connected if their intersection is nonempty. Intuitively clear and a proof of the general case can be found here: Union of connected subsets is connected if intersection is nonempty

Now to find their intersection, plug in $y = 0$ into

$$x^2 -2x + y^2 = 0 \Rightarrow x^2 - 2x = x(x-2) = 0 \Leftrightarrow x = 0,2$$

Thus we see that $(2,0)$ is in the intersection and therefore, it is nonempty.

• The "only if" in your A is incorrect .E.g, in $R$ , with $X=[0,1)$ and $Y=\{1\}.$ – DanielWainfleet Jan 27 '16 at 1:42
• You're right. I have no idea how I misread that result I cited. Luckily, the correct direction is the applicable one here. – Future Jan 27 '16 at 1:57
• I see that you fixed it. What we need is an app that checks these things for us.:) – DanielWainfleet Jan 27 '16 at 2:13

Note that $x^2-2x+y^2=(x-1)^2+y^2-1$, so $\{\langle x,y\rangle:x^2-2x+y^2=0\}$ is just the graph of the equation $(x-1)^2+y^2=1$: it’s the circle of radius $1$ centred at $\langle 1,0\rangle$. Thus, it includes the point $\langle 2,0\rangle$. Any point on the circle is connected by a circular arc to the point $\langle 2,0\rangle$, and any point on the line segment is connected to the same point by a line segment, so the whole thing is not just connected, but pathwise connected.

Here is a visualization:

The point $P$ is shared.