# Connectedness of Euclidean plane excluding the origin

Let $X= \mathbb{R}^2 \setminus \{(0,0)\}$. It is obvious that $X$ is path-connected and thus it is connected.

The definition of connectedness of a set $A$ is stated as follows: there is no pair of open sets $U_1$ and $U_2$ such that

1. $A \subset U_1 \cup U_2$;
2. $A \cap U_1 \neq \emptyset$, $A \cap U_2 \neq \emptyset$; and
3. $U_1 \cap U_2 = \emptyset$.

Then, my question is can we prove the connectedness of $X = \mathbb{R}^2 \setminus \{(0,0)\}$ directly by the definition given above?

• This is more general case. See the answer. – MAN-MADE Aug 28 '17 at 3:54
• I suppose so, but why bother? Much more interesting is to prove R^2 with a countable number of points removed is still path connected. – William Elliot Aug 28 '17 at 4:48
• Why bother? Maybe to learn something about the given definition? +1, good question. – G Tony Jacobs Aug 28 '17 at 5:34
• Yes, it is possible, but not very elegant. There are equivalent definitions of connectedness in which it is straightforward to show this space is connected. – Ittay Weiss Aug 28 '17 at 7:09
• @IttayWeiss Can you provide the equivalent definitions of connectedness you mentioned? Thank you very much. – Simon Pun Aug 28 '17 at 7:45

This definition of connectedness is more immediately suited for proving disconnectedness rather than connectedness. Basically, this definition really defines what it means to be disconnected, and then says that something is connected if it is not disconnected. This is very different than path connectedness, where the claim is that for any two points there is a connecting entity, namely a path, between the two points. Such a definition is more immediately suited to proving connectedness. If topological connectedness could be defined, equivalently, in a manner more like path connectedness, by exhibiting a connecting entity, then it would be useful for proving connectedness.

For metric spaces there is such a reformulation of connectedness. A scale on a metric space $X$ is a choice of a positive real number $R_x$ for each point $x\in X$. Given a scale, a walk is a sequence of finitely many points such that for any two adjacent points in the sequence $x,y$, either $d(x,y)\le R_x$ or $d(y,x)\le R_y$. It then holds that $X$ is connected in the topological sense if, and only if, for any two points and any scale on $X$ there exists a walk between these two points.

To obtain a similar characterisation for arbitrary topological spaces use the fact that every topological space is metrisable, provided the metric is allowed to take values in structures more general than the non-negative real numbers (and dropping the requirement that the metric is symmetric). This formalism was developed by Flagg, allowing the metric to take values in a value quantale (Quantales and continuity spaces). This approach was shown to be categorically equivalent to standard topology (A note on the metrizability of spaces). The adaptation of the notion of connectedness from classical metric spaces to Flagg's formalism was done in Metric characterisation of connectedness for topological spaces, where classical connectedness and uniform connectedness are shown to be two instances in a whole hierarchy of notions of connectedness (to which path connectedness does not belong). Among other things, some classical results about connectedness are proved using the new formalism.

Assume that there are two such open sets $U_1$, $U_2\subset\dot{\mathbb R}^2$. Then there is a point ${\bf a}\in U_1$ and a point ${\bf b}\in U_2$, and after a small relocation of ${\bf b}$ (if necessary) we may assume that the segment $$\sigma:\>[0,1]\to{\mathbb R}^2, \qquad t\mapsto(1-t){\bf a}+ t{\bf b}$$ does not hit the origin. Put $$\tau:=\sup\bigl\{t\in[0,1]\,\bigm|\,\sigma(t)\in U_1\bigr\}\ .$$ Then $0<\tau<1$, since $U_1$ is a neighborhood of ${\bf a}$ and $U_2$ is a neighborhood of ${\bf b}$. I claim that the point ${\bf c}:=\sigma(\tau)$ cannot belong to either one of $U_1$ and $U_2$.

Proof. If ${\bf c}\in U_1$ then $\sigma(\tau+\delta)\in U_1$ for some $\delta>0$, contradicting the definition of $\tau$, and if ${\bf c}\in U_2$ then $\sigma(\tau-h)\in U_2$ for all sufficiently small $h>0$, which contradicts the definition of $\tau$ as well.$\qquad\square$

It follows that there cannot be two such open sets $U_1$ and $U_2$.

• It isn't that easy to see. Since both sets are open if $c \in U$ then it is contained in an open ball $B$ within $U$. So if in $U_1$ then $B \cap \{\sigma(t)\}$ contains a point with $t \gt \tau$ contradicting $\tau$ being supremum. Conversely, if in $U_2$, then $B \cap \{\sigma(t)\}$ contains points of $U_1$ contradicting $U_1, U_2$ being disjoint. So, $\dot{\mathbb R}^2$ is not the union of two disjoint open sets and therefore connected. (I think ?) – Tom Collinge Aug 28 '17 at 14:08
• @TomCollinge: I have added the details. – Christian Blatter Aug 28 '17 at 14:38

First a proof of the connectedness of $\Bbb R^2$:

You know that $U$ must be open and closed so assume there is a $x\in U$ and since $U$ is open there is also an open ball around $x$ in $U$. Since $U$ is also closed the boundary circle is also contained in $U$. But any point on the circle has again an open neighborhood in $U$ and by compactness of the circle we know that a bigger open ball around $x$ is contained in $U$.

Now there is no bound for the balls around $x$ hence $U= \Bbb R^2$.

So how can we adopt this to the case of $\Bbb R^2$ with the origin removed? You could use this two show that two points $x,y$ which have closer distance than one oft their length lie in the same component. But by choosing such points carefully you can show that there can be just one component.