0
$\begingroup$

Show that if $U$ is open connected subspace of $\mathbb{R^2}$, then $U$ is path-connected.

You can see here but:

My Attempt:

Let $f:[a,b]\rightarrow \mathbb{R^2}$ be straight line paths joining $x,y$ such that $x,y\in U$.

Now if $f[a,b]\subset U$, we are done. If not, consider points $x_1,x_2,\dots,x_n\in U$ such that joint $x_i,x_{i+1}$ straight line will be inside $U$, also joining straight line$x,x_1$ and $x_n,y$ are in $U$. We can do this since $U$ is connected and open, then those paths must be a subset of $\text{int}\space U$. Those joining those straight line we will get a path from $x$ to $y$. Hence $U$ is path-connected.

Is this argument ok?

Solution using the hint given in the question:

Hint: Show that given $x_0\in U$, the set of points that can joined to $x_0$ by a path in $U$ is both open and closed in $U$.

My attempt:

Suppose $U_{x_0}$ is such same type of set described in hint. First note that $U_{x_0}$ is path-connected. $y\in \text{bd}\space U_{x_0}$. Now consider $U_y$ is same type of set described in hint. Then $U_y$ intersects $U_{x_0}$ since $U$ is open and connected. But since both $U_{x_0}$ and $U_y$ path-connected, hence $U_{x_0}\cup U_y$ is path-connected. Hence Hence $y\in U_{x_0}$ which implies $U_{x_0}$ is closed.

Again, $y\in U_{x_0}$. Let $V$ is a neighborhood of $y$. Then $V\subset U_y=U_{x_0}$. Hence $U_{x_0}$ is open.

Then $U_{x_0}$ is is a separation of $U$. But $U$ is connected. Hence $U=U_{x_0}$ which implies $U$ is path-connected.

Is this argument ok?

$\endgroup$
  • $\begingroup$ In your first part, it is not made clear that a finite such point sequence $x_1,x_2,\ldots,x_n$ exists $\endgroup$ – Hagen von Eitzen Sep 4 '17 at 18:32
  • $\begingroup$ Ok, that is where it was wrong. $\endgroup$ – topology_001 Sep 5 '17 at 1:43
6
$\begingroup$

No, your argument needs a lot of work.

Pick $x_0 \in U$ and use the hint to define

$$A =\{ y \in U: \exists f: [0,1] \to U \text{ continuous and } f(0) = x_0, f(1) = y\}$$

which is all points of $U$ that can be reached by a path starting from $x_0$.

Then

  • $A$ is non-empty ($x_0 \in A$ as witnesses by a constant map from $[0,1]$ with value $x_0$).

  • $A$ is open: let $y \in A$. Let $f_y:[0,1] \to U$ be the path witnessing that. Also, let $r>0$ be such that $B(y, r) \subseteq U$, where $B(y,r)$ is the standard Euclidean ball in $\mathbb{R}^n$, which is convex. This can be done as $y \in U$ and $U$ is open. Let $z \in B(y,r)$, and let $g_z$ be the straight line path from $y$ to $z$ which lies inside $B(y,r)$ by convexity. Then combining $f_y$ and $g_z$ in the usual way (i.e. $f(t) = f_y(2t)$ for $t \in [0,\frac{1}{2}]$ and $f(t) = g_z(2t-1)$ for $t \in [\frac{1}{2},1]$), we get a path inside $U$ from $x_0$ to $z$. So $z \in A$ by definition, and so $B(y,r) \subseteq A$ showing that all $y \in A$ are interior points of $A$. So $A$ is open.

  • $A$ is closed. Suppose that $y \in U$ is such that $y \in \overline{A}$. Let $B(y,r)$ be a ball around $y$ that lies inside $U$ and note that it must intersect $A$, say in a point $p \in A \cap B(y,r)$. There is a path $f_p$ from $x_0$ to $p$ and as above we combine it with a straight line path from $p$ to $y$ to show that $y \in A$. So $A$ is closed.

As $U$ is connected it only has one non-empty set that is both open and closed, namely $U$ itself. So $A = U$. Now $U$ is path-connected, if $p,q \in U$ there are paths inside $U$: $f_p$ from $x_0$ to $p$ and $f_q$ from $x_0$ to $q$. Combine the reverse of path $f_p$ with $f_q$ to create a path from $p$ to $q$ inside $U$ .

$\endgroup$
  • $\begingroup$ I have just made a correction to your answer. Do you agree to it? $\endgroup$ – Saaqib Mahmood Dec 2 '17 at 5:21
  • $\begingroup$ @SaaqibMahmuud Well caught typo! $\endgroup$ – Henno Brandsma Dec 2 '17 at 6:55
  • $\begingroup$ The justification of the ball $B(y,r)\subset U$ it's because $x\in U$ and $U$ is open. Should not be $y$ instead of $x$? $\endgroup$ – Isa Jan 6 '18 at 5:09
  • 1
    $\begingroup$ @Ella that was indeed a typo, i edited it. $\endgroup$ – Henno Brandsma Jan 6 '18 at 6:24
  • 1
    $\begingroup$ @Ella it's the standard way to make a path from $a$ to $c$ if we have paths from $a$ to $b$ and $b$ to $c$. We do the $2t$ etc. because the domain has to be $[0,1]$ in the usual definition of path connectedness. These combined path is continuous by a glueing or pasting lemma, e.g. (they conect at $t=\frac{1}{2}$). Intuitively the glueing of two paths should be clear ,this is just to formalise that. $\endgroup$ – Henno Brandsma Jan 6 '18 at 6:40
0
$\begingroup$

For this you need the chain theorem for connected sets.

If $C$ is an open cover of a connected set $S$ and $x$, $y$ in $S$, then
there is some integer $k$ and $U_1, \ldots, U_k$ contained in $C$ with $x$ in $U_1$, $y$ in $U_k$ and for all $j = 1, \ldots, k-1$, $U_j \cap U_{j+1}$
is not empty. The overlapping $U_1, \ldots, U_k$ is called a chain.

For your problem, let $U$ be a connected open subset of $\mathbb{R}^2$ or $\mathbb{R}^n$. Assume $x$, $y \in U$. $U$ is the union of a collection, $C$, of open balls.

As $C$ covers $U$, there is a finite chain from $x$ to $y$ of balls from
$C$. As those balls are overlapping and path connected, one can
easily construct a path from $x$ to $y$ through those balls.

The same argument can be used to show that any open connected
subset of a locally path connected space is path connected.

$\endgroup$
  • $\begingroup$ Thanks, but I have not read open cover, it is in the next section in the book I am reading. $\endgroup$ – topology_001 Sep 5 '17 at 4:10
  • $\begingroup$ @WilliamElliot I have just made some changes to the formatting of symbols in your post. $\endgroup$ – Saaqib Mahmood Dec 2 '17 at 5:38
  • $\begingroup$ “Need” is a strong word. “could use” fits better. $\endgroup$ – Henno Brandsma Jan 6 '18 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.