Proving/Disproving statements about connectedness and convexity

Question

I want to know if the following statements can be proven/disproven.

$1.$If $A,B \subset \mathbb{R}^n$ are connected, then $A \cap B$ are connected.

True. There are two possibilities:

1. $A\cap B=\emptyset$: then $A\cap B$ is connected.
2. $A\cap B\neq\emptyset$: then $A\cup B$ is connected, because if $f\colon A\cup B\longrightarrow\{0,1\}$ (with $\{0,1\}$ endowed with the discrete topology) is continuous then, if $p\in A\cap B$, $f(A)=\bigl\{f(p)\bigr\}=f(B)$, and therefore $f$ is constant.

$2.$ If $A,B \subset \mathbb{R}^n$ are convex, then $A \cap B$ is convex too.

True. Let $X\in A\cap B$ and $Y\in A\cap B$ which implies that $X\in A$ and $Y\in A$. Since $A$ is convex, line $XY$ lies within $A$. By similar argument line $AB$ lies within $X$. Hence line $XY$ lies within $A\cap B$.

$3.$ If $A,B \subset \mathbb{R}^n$ is connected and $A \cap B \neq \emptyset$, then $A \cup B$ is connected too.

True. Let $A = [0,1]$ and $B = (1,2)$. Then $A \cap \overline{B} = \{1\} \neq \emptyset$. However, $A \cap B = \emptyset$. So, $A \cap \overline{B} \neq \emptyset$ does not imply $A \cap B \neq \emptyset$.

$4.$ If $A,B \subset \mathbb{R}^n$ is convex and $A \cap B \neq \emptyset$, then $A \cup B$ is convex.

I think this is false. I thought of $A = \{\mathbb{R^n}, \emptyset \}$ and $B = \{\mathbb{R^n}\}$, but I don't think that's a counterexample.

Answer 1

$1$ is false. You have shown that if $A \cap B \neq \emptyset$ and $A$ and $B$ are connected, then $A \cup B$ is connected: this does not mean that $A \cap B$ is connected at all. Indeed, there are a great many counterexamples, and not even pathological ones: sticking to $\mathbb{R}^2$, take $A$ to be the unit circle minus $(0,1)$ and take $B$ to be the unit circle minus $(1,0)$. Both are clearly connected, but $A \cap B$ is the unit circle minus 2 points, which is disconnected.

$2$ is true.

$3$ is true by the proof you gave in $1$. What you've given here doesn't work, and I'm not really sure what you were trying to do with it.

$4$ is false, but I have no idea what your countexample is even supposed to mean: neither of those is a subset of $\mathbb{R}^n$. For a counterexample (again sticking with $n = 2$), consider the closed unit disks centred on $(0,0)$ and $(1,0)$: both are clearly convex, and the intersection contains $\left(\frac{1}{2},0\right)$, but the union contains both $(0,1)$ and $(1,1)$, but not $\left(\frac{1}{2},1\right)$, which lies on the line between them.

Answer 2

1. The statement is false. Take for instance, the $x$-axis in $\mathbb R^2$ and a circle centered at a point of that axis.
2. Correct.
3. True, but you have to prove it.
4. It is false. Take the $x$-axis and the $y$-axis in $\mathbb R^2$.

Answer 3

1. What happens if you intersect the symbol $\bigcup$ with the symbol $\bigcap$ ?

2. Ok.

3. This claim is actually true.

4. I don't understand what $\{\Bbb R^n,\emptyset\}$ has to do with convexity. Anyways, look at the symbol $+$.

Answer 4

If $A$ and $B$ are connected then $A\cap B$ may or may not be connected. For example in $R^2$ consider a $U$ shape region and a rectangular region cutting through the two teeth of it. The intersection has two disjoint components so it is not connected.

The convexity of the intersection is true. For the union of two sets neither the connectedness nor the convexity is necessarily preserved.

You need to reconsider your proofs.