How to show $I^2=[0,1]×[0,1]$ remove three points is connected ?
I am sure $X=I^2\backslash\{a,b,c\}$ is path connected hence connected for any $a,b,c \in I^2$, but I am not sure how to find the actual continuous path for any two points in $X$. Or is there any other method I can use to show $X$ is connected? Thanks a lot!
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You can use the fact that if there exists a punctured neighborhood $\Omega_p\subseteq C$ of a point $p\in C$ then the set $C\setminus\{p\}$ is connected if $C$ is connected.
To see how this helps let's assume that you picked away $p_1, ... p_{n-1}$ from $I^2$ without disconnecting it and we have a distinct $p_n\in I^2$ (if it's not distinct removing it will do no difference), we know that for a punctured disc of radius $r$ we have that $D_{p_n;r} \cap I^2$ is connected. Now we can choose $r$ so that the punctured disc doesn't cover any of the other points and thereby $D_{p_n;r} \cap I^2 \subseteq I^2\setminus \{p_1, ... p_{n-1}\}$.
Now for the proof of the lemma that's used. Assume that $C\setminus\{p\}$ is not connected, then we have to open disjoint sets $U$ and $V$ such that $C\setminus\{p\} = U \cup V$, but now $\Omega_p\cap V$ and $\Omega_p\cap U$ are open and disjoint sets and their union is $\Omega_p\cap(U\cup V) = \Omega_p\cap(C\setminus\{p\}) = \Omega_p$, but that contradicts the fact that $\Omega_p$ is connected.
You don't need to actually find the path - if it's pathconnected then it's connected, no matter what paths are used.
However, you could do the following: given any two points, draw a straight line between them. If that line is invalid because one of the $a, b, c$ is on it, then take a detour by using two straight lines through some corner point instead. (There is a valid detour: there are infinitely many possible detours, but each removed point invalidates precisely one detour if you take the corner point to be equidistant from the two points.)