MHB Is Path-Connectivity Equivalent to Connectivity in $\mathbb{R}^2$?

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Hey! :o

Can a set in $\mathbb{R}^2$ be path-connected only when it is connected, i.e. when we know that a set is not connected then it cannot be path-connected? (Wondering)

We have the sets
  • $\displaystyle{A=\{x\in \mathbb{R}^2 : 1\leq x_1^2+x_2^2\leq 4\}}$
  • $\displaystyle{B=\{x\in \mathbb{R}^2 : x_1^2-x_2^2=1, x_1, x_2>0\}}$

I have shown that these sets are connected. Could you give me a hint how we could check whether they are path-connected or not? (Wondering) I have also an other question. Suppose we have the set $\displaystyle{C=\{x\in \mathbb{R}^2 : x_2\cos x_1=\sin x_1\}}$. This is equivalent to $\displaystyle{C=\{x\in \mathbb{R}^2 : x_2=\tan x_1\}}$.
We have that the tangens function is not continuous on the whole $\mathbb{R}$, since it is not defined everywhere. Is this enough to say that this implies that the set is not connected? Or do we need more information to get that conclusion? (Wondering)
 
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mathmari said:
Hey! :o

Can a set in $\mathbb{R}^2$ be path-connected only when it is connected, i.e. when we know that a set is not connected then it cannot be path-connected? (Wondering)

We have the sets
  • $\displaystyle{A=\{x\in \mathbb{R}^2 : 1\leq x_1^2+x_2^2\leq 4\}}$
  • $\displaystyle{B=\{x\in \mathbb{R}^2 : x_1^2-x_2^2=1, x_1, x_2>0\}}$

I have shown that these sets are connected. Could you give me a hint how we could check whether they are path-connected or not? (Wondering)

For the first set: Make a sketch, then use the gluing lemma.
For the second set: $x_1$ is a continuous function of $x_2$ on the interval $(0,\infty)$.

mathmari said:
I have also an other question. Suppose we have the set $\displaystyle{C=\{x\in \mathbb{R}^2 : x_2\cos x_1=\sin x_1\}}$. This is equivalent to $\displaystyle{C=\{x\in \mathbb{R}^2 : x_2=\tan x_1\}}$. We have that the tangens function is not continuous on the whole $\mathbb{R}$, since it is not defined everywhere. Is this enough to say that this implies that the set is not connected? Or do we need more information to get that conclusion? (Wondering)

Could you write $C$ as the disjoint union of non-empty, closed subsets of $\mathbb{R}^2$?
 
Krylov said:
Could you write $C$ as the disjoint union of non-empty, closed subsets of $\mathbb{R}^2$?

I don't really have an idea about that. Could you give me a hint? (Wondering)
 
Krylov said:
For the first set: Make a sketch, then use the gluing lemma.
For the second set: $x_1$ is a continuous function of $x_2$ on the interval $(0,\infty)$.

For the first set: Can we see that only with the graph? (Wondering)

For the second set: We have that $x_1=\sqrt{1+x_2^2}$. What do we get from that? (Wondering)
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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