Graph theory simple question but confused

For pic related I wrote:

" If C(u) and C(v) do not share any vertex then the result is true, so suppose C(u) and C(v) share a vertex which we will call 's'. This implies that there exists a path from u in C(u) to v in C(v) and is strongly connected as C(u) and C(v) were strongly connected components, this can be seen as: {u, u2} {u2, u3} {u3, u_n} {u_n, s} {s, v_n} {v_n, v3} {v3, v2} {v2, v}. But since C(u) and C(v) share the common vertex 's', they are no longer components as its against the definition of component, so it follows that C(u) and C(v) do not share any vertices."

But my professor said this isn't right, (he specifically just said "no" in an email, nothing else)

I've been thinking about whats wrong with what im saying and I can't see anything wrong with it, or even if conceptually what im doing is wrong what other way could I show that C(u) and C(v) do not show a vertex.

I know there is some ambiguity in graph theory as different people consider different things but could I get a general idea on how to prove this problem?

friendly bump

I think the problem is with your conclusion that C(u) and C(v) are no longer components.

I can't help with the question, but this is the bateleur of knowledge, think upon things you know that once you did not, now think upon things you do not know and how you will come to know them and look back at the present in the same way, hopefully this will motivate you

I coulda added somewhere "But in our hypothesis we considered C(u) and C(v) to be components, thus by contradiction C(u) and C(v) cannot share verticies"

I do agree that it could be the reason why he said no, however I don't think that anyone is that cold hearted that they couldn't point out such a small thing (technically I didn't even need to point it out since its implied)

I think im missing something important else I wouldnt have gotten the "no" cold reply

thanks bateleur of knaaaledge

>>But in our hypothesis we considered C(u) and C(v) to be components, thus by contradiction C(u) and C(v) cannot share verticies

Why is this true?

Well we defined C(u) and C(v) in a way that they are strongly connected... so since I showed that if they did share a vertex they cannot be considered components by definition, which is the contradiction, I defined C(u) and C(v) to be components yet somewhere along my working they turned out not to be components

I don't see how you're getting that they are no longer components. I think the conclusion that you want to make is that they have to be the same component. And this is what you're trying to show.

Suppose you have a strongly connected graph, with more than one vertex. How many strongly connected components does this graph have?

Further, if u and v are distinct vertices, what then is the relationship between C(u) and C(v)?

Well the case we have to prove is that when C(u) and C(v) share a vertex, but that can't even be possible since then then "C"(u) and "C"(v) are connected, but if theyre connected then how are they cant be separate components to begin with..?

>Suppose you have a strongly connected graph, with more than one vertex. How many strongly connected components does this graph have?
Just the 1, itself?

>Further, if u and v are distinct vertices, what then is the relationship between C(u) and C(v)?
Well if u and v are in the set of verticies and are distinct then they wouldnt have a relationship since they cant be the same component?

Why are you assuming they are separate components?

Well the question says that C(u) is a component which is a subset of G and C(v) is a component which also a subset of G. So either they are different or they are the same thing, if theyre the same thing C(u) and C(v) will share every single vertex since theyre the same thing, which doesnt make sense, so they need to be separate components to begin with, not to mention u, v are also distinct so they cannot be the same component

>not to mention u, v are also distinct so they cannot be the same component
No

Can you explain? My current understanding of the word 'distinct' means they cannot be the same thing, if two things are distinct theyre different.

If u and v are distinct but are the same points on the same component how are they "distinct" when theyre the same thing?

If s is in C(u) and in C(v), then there exists a path from u to c and back, and there exists a path from v to c and back. Let the first pair of paths be p1 and p2, so u -- p1 -- c and c -- p2 -- u are valid paths. Likewise let the second pair of paths be p3 and p4, such that v -- p3 -- c and c -- p4 -- v are valid paths.

I claim u and v are strongly connected, in that there exists a pair of paths p5 and p6 such that u -- p5 -- v and v -- p6 -- u are valid. Specifically, p5 is (u) -- p1 -- c -- p4 -- (v) and p6 is (v) -- p3 -- c -- p2 -- (u). Thus v is in C(u) and u is in C(v).

But this same proof applies for any u' in C(u) and any v' in C(v). So every vertex in C(u) is in C(v) and vice versa. So C(u) = C(v).

If s is in C(u), then C(u) = C(s).

How many distinct vertices does this graph have? How many distinct components does it have?

8 distinct verticies, 3 distinct components? assuming theyre all a subset of some graph "G"

This is basically what I wrote except for
>Thus v is in C(u) and u is in C(v).
How can this be though? u,v are DISTINCT.... can distinct things be the same thing??? I don't understand how they can say one thing then mean another thing, its either the same thing or its not?

Notice that the vertices labeled 6 and 5 belong to the same component. And these vertices are distinct. But the component is the same.

u and v are distinct; C(u) and C(v) are not necessarily distinct (and in this case, they are not distinct). U and v are distinct elements in a shared strongly-connected component. By way of analogy, 2 and -2 are distinct numbers, but if f(x) = x^2, f(2) = f(-2).

ohhhh, I understand now, thanks a lot for the help guys