Pathwise connected

Give an example: Let $A$ and $B$ be pathwise-connected subsets of $\mathbb{R}^2$, whose intersection $A?B$ is nonempty. The union $A?B$ is pathwise connected. 

*Use the fact that a convex subset is pathwise connected for one of the subsets. 

Example:
$A=\{(x,y)\in \mathbb{R}^2 | \hspace{1mm} |x|+|y| \leq 10\}$
$B=\{(x,y)\in \mathbb{R}^2 | \hspace{1mm} x^2+y^2=25\}$
and let $C=A \cup B$. Define the function $f:C\rightarrow \mathbb{R}$ by
$f(x,y)=\frac{5x+2y}{{\sqrt{x^2+y^2}}+1}$ 

Something similar to this example. But I need different subsets and function from this.

  • Are you sure you wrote the question correctly?

  • Which part is wrong?

  • Well, you could take A = B = R^2. All sets are convex and path-connected, A intersection B = R^2 is non empty and A union B = R^2 which is again convex and path-connected. One could also find easy examples with A different to B.

  • I have edited the question to make it clearer.

  • Please fix: " Let A and B of be pathwise-connected of \mathbb{R}^2". That statement is not correct.

  • Sorry for the mistake.

  • The bounty is low.

  • You just need an example or you also need a proof that it satisfied the conditions?

  • A simple proof to show that each two subsets are pathwise connected or just hint would be fine.

  • I mean each subsets

  • I'm confused now. Do you need to prove that, "If A and B are path connected with non-empty intersection, THEN their union is also path connected"? Or do you need an example of this behavior? If it's just the example, I can answer that very quickly.

  • Just the example with explanation.

Answer

Answers can be viewed only if
  1. The questioner was satisfied and accepted the answer, or
  2. The answer was disputed, but the judge evaluated it as 100% correct.
View the answer

1 Attachment

  • Sorry can I request for a more complicated subsets? Just like the example I have given. I wish to have something like x^2+y^2=25 (e.g. a shape)

  • You can take A= {(x,y)| y>=x^2}, this is everything above the parabola y = x^2 and it is path connected. Take B = {(x,y)| y<=1-x^2}, this is everything below the parabola y = 1- x^2 . The intersection is not empty and their union is path connected.

  • Or can you help me to verify if I can define the two subsets as: A=[-3,3]x[-4,4] which is a generalised triangle and B={(x,y) | x^2+y^2=4}.

  • May I have some steps to show that y>=x^2 by using the definition of pathwise connected?

  • I also need a function with AUB as its domain, as shown in the example :)

  • The function sin(x+y) is defined on AUB.

  • The definition of path connected means that every pair of points in the set can be joined by a path. In this case, a line between two points will join them.

  • Can I have a function that is more complicated?

  • Sin((5x^2-2y^3)/(1+^x4+y^6))

  • Thank you!

  • Is each of these subsets is pathwise-connected? A=[-3,3]x[-4,4] which is a generalised triangle and B={(x,y) | x^2+y^2=4}

  • A=[-3,3]x[-4,4] is just a square, but yeah, they are path-connected.

  • Thank you for your help!

The answer is accepted.