Locally connected for sub space topology of $\Bbb R^2$ Consider the following subspace $X:=\Bbb R^2\setminus(\Bbb Q\times \Bbb Q)$ of $\Bbb R^2$, where $\Bbb R^2$ with the usual topology. I would like to check this space in the terms of various kinds of connectedness.

$X$ is a path connected by using the fact says "$\Bbb R^2\setminus K$ is path-connected for any countable set $K$" as $\Bbb Q\times \Bbb Q$ is countable. In  particular, $X$ is connected. Is that right?


Next, I would like to check locally connected but I found it hard by using the definition. Any idea?

 A: Your argument as to the path-connectedness of $\Bbb R^2\setminus \Bbb Q^2$ is correct, if that fact is indeed allowed to be used (i.e. is part of your curriculum or an earlier exercise).
One local base of neighbourhoods of a point $(x,y) \in X$ is sets of the form $(x-r,x+r) \times (y-r,y+r)\setminus \Bbb Q^2$ for $r>0$ (just a local base at $(x,y)$ intersected with the subspace $X$. As open intervals are homeomorphic to $\Bbb R$, the same fact applies and $X$ is locally path-connected as well.
A: Looks right. If you have two points in $R^2-K$ say $a$ and $b$ then one can draw uncountable many circles passing through these two points. Since each such circle is determined by its radius $r$ and any (sufficiently large) value of $r$ would generate different arc. Since $K$ is countable there is an arc without points from $K$ i.e. contained in $R^2-K$.
The second part is kind of similar though here we use, kind of local path connectness. That is if a disk $D$ (in $R^2-K$) is not connected then it could be represented as a sum of two non empty, disjoint open sets, say $A$ and $B$. But then if $a\in A$ and $b\in B$ by the argument as above, there is arc joining $a$ and $b$ contained in $D$. In words, we have an arc which can be represented as a sum of two disjoint open sets (i.e. $arc\cap A$ and $arc\cap B$) which is not possible.
