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Previously I asked a question about path connected sets specifically dealing with the closure of a set. I'm still trying to grasp the topic and have another example I was looking to get some feedback on.

Definition for path connected: A topological space $X$ is path connected if for every $x,y \in X$ there is a path in $X$ from $x$ to $y$. A subset $A$ of a topological space $X$ is path connected in $X$ if $A$ is path connected in the subspace topology that $A$ inherits from $X$.

Now, Consider $\mathbb{R}^2$ with the standard topology and define two sets $A$ and $B$ on $\mathbb{R}^2$ as:

$A \subset \mathbb{R}^2$ given as $A=\{(x,y)\ s.t.\ 0\le x \le 5$ and $y \le x \}$
$B \subset \mathbb{R}^2$ given as $B=\{(x,y)\ s.t.\ 0\le x \le 5$ and $x \le y \}$
Both A and B are path connected.

I believe the following sets are path connected: $int(A) \cup int(B)$, $A \cup int(B)$, and $A \cup B$. Any feedback on these three sets? Thanks.

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P Haggerty
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1 Answers1

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All of the relevant properties fall out once you notice that $int(A)=\{(x,y):0<x<5 $ and $y<x\}$ and $int(A)=\{(x,y):0<x<5 $ and $y>x\}$.

So $int(A)\cup int(B)$ is disconnected as the line $y=x$ separates them, while $int(A)\cup B$ and $A\cup int(B)$ are still path connected, as they are the square with boundary edges removed.

user402287
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  • Then $A \cup B$ would be path connected as would be $A \cap B$, but $int(A) \cap B$ would not be path connected as the intersection is $\emptyset$, right? – P Haggerty Dec 29 '16 at 01:24
  • @PHaggerty Your intuition is correct. However, from the definition above, $\emptyset$ would be vacuously path connected, since to be not path connected, you need two distinct points which cannot be connected by a path, and you cannot find two distinct points in $\emptyset$. – user402287 Dec 29 '16 at 01:48