Thread: Connected Subset

Let A be a connected subset of . Suppose that the points (0,0,1) and (4,3,0) are in A.

a. Prove that there is a point in A whose second component is 2.
b. Prove that there is a point in A whose norm is 4.

a. Let u = (0,0,1) v = (4,3,0) be points in A. Next, I find the symmetric equation of the line l that contains the points u,v.




Thus,.


We use the first point u to find the symmetric equation of l.
Thus, the equation yields:





It is sufficient to show that there exists a y such that y/3 = 2 that disconnects A into two disjoint sets?


b. My question is could I just take the norm of the points u = (0,0,1) and v = (4,3,0) which is and that there must exist a a point whose norm is 4 because it is less than ?


Thank you for your time.

a) My immediate take on this (long time since I've done topology) is as follows.

Since A is connected, there exists a continuous path from u to v.

Thus (I expect there's a theorem somewhere that proves this), the projection of this path onto each of the 3 co-ordinates is also continuous.

That is, the x-coordinate of that path must be continuous from 0 to 4, the y-coordinate from 0 to 3 and the z-coordinate from 1 to 0.

As the y-coordinate is continuous from 0 to 3, it must pass through 2 somewhere along it.

b) I'll have to pass on that one as I can't remember what a norm is.

Hello

Originally Posted by Matt Westwood

Since A is connected, there exists a continuous path from u to v.

This is not necessarily true. (it'd be true if A were path-connected)

Originally Posted by Paperwings

Let A be a connected subset of . Suppose that the points (0,0,1) and (4,3,0) are in A.

a. Prove that there is a point in A whose second component is 2. (...)

a. Let u = (0,0,1) v = (4,3,0) be points in A. Next, I find the symmetric equation of the line l that contains the points u,v.




Thus,.


We use the first point u to find the symmetric equation of l.
Thus, the equation yields:





It is sufficient to show that there exists a y such that y/3 = 2 that disconnects A into two disjoint sets?

If you find two disjoint, proper subsets of A which are open in A, yes, A is disconnected... but how can you find two such sets given that A is, by definition, connected ?

This idea of disconnecting A is however useful if you use contradiction :

Assume that there is no point in A whose second coordinate is 2. This means that the intersection of A with the plane whose equation is y=2 is the empty set. Now, can you find two disjoint, proper subsets of A which are open in A ? (this would give us a contradiction : A would have to be both connected and disconnected)

Hint (highlight to read) : *Consider the intersection of A with the half space { (x,y,z) in R^3 | y<2 } and the intersection of A with the half space { (x,y,z) in R^3 | y>2 }.*

b. Prove that there is a point in A whose norm is 4. (...)

b. My question is could I just take the norm of the points u = (0,0,1) and v = (4,3,0) which is and that there must exist a a point whose norm is 4 because it is less than ?

No, the word "because" hides too many things in this sentence. Once again, you can try using contradiction : assume that there is no vector in A whose norm is 4. (the idea is the same as for the first question)

Assume that there is no point in A whose second coordinate is 2. This means that the intersection of A with the plane whose equation is y=2 is the empty set. Now, can you find two disjoint, proper subsets of A which are open in A ? (this would give us a contradiction : A would have to be both connected and disconnected)

Thanks flyingsquirrel.

So, assume that for all x,z, . I need to show that the plane
into two open disjoint, proper sets.

The first set is (0,0,1) since the y component is 0 < 2 which belongs in the plane { (x,y,z) in R^3 | y<2 }
The second set is (4,3,0) since the y component is 3 > 2 which belongs in the plane { (x,y,z) in R^3 | y>2 }

Similarly, this can be done with the norm. There is point in A such that {a, ||a|| = 4}

The first set is the norm of (0,0,1), which is 1.
The second set is the norm of (4,3,0) which is 5.

Which contradicts A being connected.

I have a good hunch this is correct.

Originally Posted by Paperwings

(...) I have a good hunch this is correct.

No, it's not correct.

So, assume that for all x,z, . I need to show that the plane
into two open disjoint, proper sets.

My mistake : I forgot to say that the union of these two subsets has to be A. (the hint I gave remains correct though)

The first set is (0,0,1) since the y component is 0 < 2 which belongs in the plane { (x,y,z) in R^3 | y<2 }

No, there can be more than one point in the intersection of A with {(x,y,z) in R^3 such that y>2}. Plus you didn't check that this set is open*, that it is a proper subset of A... I wrote the answer for this question (see attached picture) hoping it'll help you. As the proof for the second question is similar, I let you write it on your own. (and let me know if something is not clear in the answer to the first question)

(*) which it is not since it's a singleton of R^3 with the Euclidean topology...

Originally Posted by JaneBennet

It is true in . For topological spaces in general, connectedness does not imply path-connectedness – but in the case of the two concepts happen to be equivalent.

I searched the web for this and the only thing I found is that "In dimensional space (...) every open connected set is path connected." (comes from the paragraph titled " Connected plus Locally Path Connected Implies Path Connected " of this page) If this is what you meant, what happens if the set is not open ? If that's not what you meant, may I ask you to give me a link to a webpage where I could read a proof of the result you're talking about?

Since is (path-)connected (see point above) every point on the straight line joining two points in lies in .

Either I'm missing something either you're saying that if a subset of is path-connected then it is convex. That seems wrong : for example the union of the planes and is path-connected since and are both path connected and . However, isn't convex : for and , the point lies in the line segment but is neither in nor in .