Let $P = \{(x,y) \in \mathbb R^2 : x^2+y^2<1 \}$ and $Q = \{(1,x) : x \in \mathbb R\}$. Is $X=P \cup Q$ connected?

If I look just at $P$, I conclude $P$ is connected. $Q$ is a vertical line and is also connected. How to see their union is connected or not?

I can see that $X=P \cup Q$ is pathwise connected (geometrically) if I can join any two points of $X$ by a path (line) which lies inside this set, does pathwise connectedness implies connectedness?

Observe that $P$ (unit open disc) and $Q$ (vertical line touching $P$ at $(1,0)$) are both convex, so path-connected.

It's clear that the segment $S = [\frac12, 1] \times \{0\}$ is contained in $P \cup Q$.

• $[\frac12, 1) \times \{0\} \subseteq P$
• $(1,0) \in Q$

The segment $S$ is path connected, and both path connected $P$ and $Q$ have nonempty intersection with $S$, so $P \cup S \cup Q = P \cup Q$ is path connected. (Edited in response to MikeMathMan's comment)

To show the path connectedness implies connectedness, we use a classic proof by contradiction. Let $X$ be a path-connected space and $U$, $V$ be two open sets that separate $X$. (i.e. $U \sqcup V = X$, where $\sqcup$ denotes disjoint union.) $\gamma: [0,1] \to X$ be a continuous path starting at $u \in U$ and ending at $v \in V$. (i.e. $\gamma(0) = u$, $\gamma(1) = v$) Due to the continuity of $\gamma$, $\gamma^{-1}(U)$ and $\gamma^{-1}(V)$ are both nonempty and open in $[0,1]$, but they separate $[0,1]$, contradiction.

• Thank you , you apply this theorem ? - Union of P and Q are connected if they have non empty intesections provided P and Q are connected , so you constructed one more connected set S so that S,P,Q are connected and S intersection P intersection Q is non empty am i right ?? Apr 6 '18 at 20:34
• @user534210 Yes, it's an intuitive result "the union of a collection of connected subspaces of $X$ that have a point in common is connected" (Thm 23.3 in Munkres' Topology.) I've edited my answer to give a stronger and more intuitive (in the visual sense) result. Apr 6 '18 at 20:43
• Thank you very much .... Apr 6 '18 at 20:49
• @user534210 I've added a picture to give an illustration. If "connectedness" is replaced by "path connectedness" in the quoted theorem, then it's much easier to see. Apr 6 '18 at 20:53
• I think you 'classic proof' needs a little work. Uh, for your '... and $\gamma$ be a continuous path' - is the constant path OK? Apr 7 '18 at 14:08

Yes, the union is path connected, thus it is connected.

Note that if you have two points in the union you can always connect them with a path.

In case that both points are on the line or both are in the disk the path is just a segment connecting the points.

In case that one is inside the disk and the other on the line you connect the points to the point (1,0) and the path is the union of two segments.

The OP's headline question wants to know if a subspace of $\mathbb R^2$ is connected, but then in the detailed posts asks

... does pathwise connectedness implies connectedness?

But they also state

... I can join any two points of X by a path (line) which lies inside this set ...

Here we will solve the problem directly, in term of 'pure' point-set topology; it won't be necessary to use the concept of a function.

To define the topological product of two spaces we specify a 'cartesian product' base. For $\mathbb R^2$ it is not difficult to show that a different base can be chosen that generates the result - we can use open disks.

Exercise: Show that all lines and line segments (can be either open or closed at an endpoint) in $\mathbb R^2$ are connected.

Let $(L_\alpha)$ be the family of all lines passing through the point $(1,0)$. Let

$\tag 1 S_\alpha = L_\alpha \bigcap X$

It is easy to see that each $S_\alpha$ is a line ($x = 1$) or a line segment. Moreover,

$\tag 2 \bigcup S_\alpha =X$ and $\tag 3 \bigcap S_\alpha = \{(1,0)\}$

But a well known result allows us to now conclude that $X$ is connected.