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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?

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  • $\begingroup$ In your first part, it is not made clear that a finite such point sequence $x_1,x_2,\ldots,x_n$ exists $\endgroup$ Commented Sep 4, 2017 at 18:32
  • $\begingroup$ Ok, that is where it was wrong. $\endgroup$ Commented Sep 5, 2017 at 1:43

2 Answers 2

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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 $U$ .

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  • $\begingroup$ I have just made a correction to your answer. Do you agree to it? $\endgroup$ Commented Dec 2, 2017 at 5:21
  • $\begingroup$ @SaaqibMahmuud Well caught typo! $\endgroup$ Commented Dec 2, 2017 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 $x$? $\endgroup$
    – user486983
    Commented Jan 6, 2018 at 5:09
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    $\begingroup$ @Ella that was indeed a typo, i edited it. $\endgroup$ Commented Jan 6, 2018 at 6:24
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    $\begingroup$ @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. $\endgroup$ Commented Jan 6, 2018 at 6:40
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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.

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  • $\begingroup$ Thanks, but I have not read open cover, it is in the next section in the book I am reading. $\endgroup$ Commented Sep 5, 2017 at 4:10
  • $\begingroup$ @WilliamElliot I have just made some changes to the formatting of symbols in your post. $\endgroup$ Commented Dec 2, 2017 at 5:38
  • $\begingroup$ “Need” is a strong word. “could use” fits better. $\endgroup$ Commented Jan 6, 2018 at 7:39

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