# The graph is connected

1. Oct 3, 2012

### happysauce

1. The problem statement, all variables and given/known data
Let f:X→Y we a continuous map. Let
Γ(x)={(x,f(x))∈X×Y}

a) Show that if X is connected, then Γ is connected.
b) Show that if Y is hausdorrf then Γ is a closed subset of X×Y.

2. Relevant equations
thm: if X is connected the image of a continuous map f:X→Y is connected.
thm: The product space of two connected spaces is connected.
thm: X is hausdorrf iff the diagonal is closed (?)
def: A space is hausdorrf if each pair of distinct points have neighborhoods that are disjoint.

3. The attempt at a solution

a) Since f:X→Y is continuous, the image is going to be connected. Therefore image(f(X))$\subset$ Y is connected. Also given two connected spaces the product space is also connected, therefore X×image(f(X)) is connected. Therefore Γ is connected? This seemed too easy so im not sure if I left a huge hole here...

b) Y is hausdorrf. Therefore given any points y1≠y2 we can find a disjoint neighborhoods U and V around y1 and y2 respectfully. I also know that the diagonal, D, is closed, {(y,y)|y is in Y}. Not sure what to do next. Maybe consider the complement and try to do a proof similar to the diagonal proof? The complement of D would be open. I dont think Y being hausdorrf implies that f(x1) and f(x2) have disjoint neighborhoods so I'm not sure where to go next.

Edit: *Being Hausdorrf implies that sequences converge to at most one point. *Does this mean that f(x_n) converges to f(x) in this case? *

I don't want any flat out answers, I just want to know what I should be looking for.
Also new here to the physics forums, if my write up is awful please tell me :)

Last edited: Oct 3, 2012
2. Oct 3, 2012

### happysauce

I might have it. consider y≠f(x) then (y,f(x)) is in the complement of Γ. Take y in U and f(x) in V. Since Y is hausdorrf x and f(x) have disjoint nighborhoods in Y. Suppose U$\cap$ V ≠∅ then there exists x=f(x), which is a contradiction, therefore given any ordered pair (y,f(x)) the neighborhood UxV is open in the complement of the graph? Therefore the graph is closed?

3. Oct 3, 2012

### jbunniii

How did you conclude that? It's certainly not true that an arbitrary subset of a connected set is connected.

Try working with the definition. Assume $\Gamma(X)$ is disconnected. Then $\Gamma(X)$ is the disjoint union of two open sets $U$ and $V$. What can you say about the inverse images, $\Gamma^{-1}(U)$ and $\Gamma^{-1}(V)$, of these sets?

4. Oct 3, 2012

### happysauce

I concluded this with a theorem from my book. "The image of a connected space under a continuous map is connected." So I thought maybe given X is connected the image of f was connected, and I just threw in the subset of Y because every point in the graph is an element in XxY.