Removal of number of distinct points which doesn't change the connectedness of the set Let $$X=\{(x,y)\in\Bbb R^2:x^2+y^2=1\}\cup([-1,1]\times\{0\})\cup(\{0\}\times[-1,1])$$
Let $n_0=max\{k:k\lt\infty\}$ there are k distinct points $p_1,p_2,...,p_k\in X$ such that $X-\{p_1,p_2,...,p_k\}$ is connected.
What is the value of $n_0$ ?
I plotted the graph of the given set and found that the given set is a circle, with centre $(0,0)$ radius $1$ in the plane $\Bbb R^2$ with diameter in $x-$axis and $y-$axis. Now, I am unable to proceed further so help is needed. Thank you.
 A: Consider $X$ as the union of the unit circle $S^1$ and the "unit cross" $Y = [-1,1] \times \{0\} \cup \{0\} \times [-1,1]$. Now observe that whenever you remove a point from $Y$ the resulting space is disconnected. On the other hand, you can remove exactly one point from $S^1$ and still have a connected space.
Now, if you remove the centre of the cross, then you can remove only one other point (on $S^1 \setminus Y$) and still have a connected space.
On the other hand, if you remove one point from three of the "branches" of the cross $Y$, then you still have a connected space, and you can still remove one point from $S^1 \setminus Y$ or from the vertex of the remaining branch. Since the resulting space is contractible, though, you cannot remove any other point.
Due to the symmetry of $X$ it isn't hard to see that this is the maximal case, so the answer to your question is $4$.

Basically, the idea is that $X$ contains five loops: the four "holes" and the outer circle. You can break all of those loops and then you can still reach any point with a path starting from some fixed, given point, but you can't if you remove any more points. This can be formalized with the notion of fundamental group.
The answer then follows because $4$ is the maximal number of points that you need to remove to break those loops, because the first point you remove will break at least two loops.
