Let $A = B^3((1,0,0),1) \cup B^3((-1,0,0),1)$ Show that $A$ is disconnected.

Question

Let $A = B^3((1,0,0),1) \cup B^3((-1,0,0),1)$ and $E = \overline{B^2}((1,0,0),1) \cup \overline{B^2}((-1,0,0),1)$. Show that $A$ is disconnected and that $E$ is connected.

To show that $A$ is disconnected assume the contrary. This would mean that $$B^3((1,0,0),1) \cap B^3((-1,0,0),1) \ne \emptyset$$ which I know is a contradiction since the balls are open, but I'm not sure how to show this?

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Similarly to show that $E$ is connected assume that it's not. This would imply that

$$\overline{B^2}((1,0,0),1) \cap \overline{B^2}((-1,0,0),1) = \emptyset,$$

which is a contradiction since

$$0 \in \overline{B^2}((1,0,0),1) \cap \overline{B^2}((-1,0,0),1),$$

but I'm not sure how to show this formally?

Answer 1

To show that $A$ is disconnected assume the contrary. This would mean that $B^3((1,0,0),1) \cap B^3((-1,0,0),1) \ne \emptyset$

Not true. For example, $[0,1)\cup [1, 2)$ is a connected set, but your argument could be used to prove that it is disconnected.

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To prove that a set $X$ is disconnected, you must find $A,B\subset X$ such that $A$ and $B$ are both open (or both closed) in $X$, and you have $A\cup B=X, A\cap B=\emptyset$. Hint: In your case, the obvious choice of how to split your set into two disjoint sets is also the correct one.

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Similarly to show that $E$ is connected assume that it's not. This would imply that $\overline{B^2}((1,0,0),1) \cap \overline{B^2}((-1,0,0),1) = \emptyset$

Again, not true.

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To prove $E$ is connected, it is easiest to show that $E$ is path connected.

Answer 2

$A$ is a union of two disjoint non-empty open subsets, $B^3((1,0,0),1)$ and $B^3((-1,0,0),1)$ and so by definition is disconnected.

$E$ is the union of two intersecting (in $(0,0,0)$) connected sets $\overline{B}^3((1,0,0),1)$ and $\overline{B}^3((-1,0,0),1)$ and as such is connected by a standard theorem on connected subsets.