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

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Sets
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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)
 
Physics news on Phys.org
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)
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
Back
Top