# If $A$ countable then $\Bbb{R}^2\setminus A$ is path connected [duplicate]

Let $A\subset\Bbb{R}^2$ be countable. I need to prove that $\Bbb{R}^2\setminus A$ is path connected.

I know that through each of $\Bbb{R}^2\setminus A$, there pare uncountably many straight lines, and as there are only countably many points in $A$, uncountably many of these lines will not contain any point of $A$. But why do I construct a path between any two points.

Also can this result be generalised, so that:

If $X$ is uncountable and $A$ is a countable subset of $X^2$, then should $X^2\setminus A$ be path connected.? (where $X$ and $X^2$ are path connected of course)

## marked as duplicate by GNUSupporter 8964民主女神 地下教會, Brahadeesh, Arnaud D., supinf, José Carlos Santos general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 1 at 13:59

• I just can't imagine if $A$ is the set of all rational coordinates, then how one does the continuous path between, say, $(\pi,\pi)$ and $(e,e)$. – user284331 Dec 9 '17 at 6:35
• Exactly, but this is a problem from the book on topology by Munkres – Abishanka Saha Dec 9 '17 at 6:36
• Hint: For $P,Q\in \mathbb{R}^2\backslash A$, consider the straight lines through $P$ and lines through $Q$. – Phil. Z Dec 9 '17 at 6:37
• math.stackexchange.com/questions/155952/… – user284331 Dec 9 '17 at 6:38
• @Phil.Z Ok, so there must be a line through $P$ and another line through $Q$ which intersect. Right? – Abishanka Saha Dec 9 '17 at 6:41

Let a,b be two points on the plain and not in A.
Draw an arc of a circle of radius r with ab as a cord.
There are uncountable many such arcs and as they are pairwise disjoint execept at the endpoints, almost all of them will miss A. Thus a circlular arc from a to b missing A.

• There are only two semicircles with a given diameter. Now, if you were talking about circular arcs.... – Lord Shark the Unknown Dec 9 '17 at 6:56
• @LordSharktheUnknown. r is not fixed. – William Elliot Dec 9 '17 at 11:28
• The radius of a circle is always half its diameter. – Lord Shark the Unknown Dec 9 '17 at 11:34
• I was going to post a complicated construction but after seeing this, I won't...............+1 – DanielWainfleet May 12 '18 at 14:47

Recall that a countable set can only be partitioned into a countable family of blocks. This does not require the Axiom of Choice (just saying).

Exercise 1: Show that there is a line of $\Bbb{R}^2$ wholly contained in $\Bbb{R}^2\setminus A$.

Exercise 2: Choose a line $L_0 \subset \Bbb{R}^2\setminus A$. Show that any point $P \in \Bbb{R}^2\setminus A$ with $P \notin L_0$ belongs to a line $L_P$ in $\Bbb{R}^2$ such $L_P \subset \Bbb{R}^2\setminus A$ and $L_P \cap L_0 \ne \emptyset$.