# Connected Hausdorff spaces and graphs

• Mar 11th 2010, 04:29 PM
Drexel28
Connected Hausdorff spaces and graphs
I came up with the following question, and would appreciate if I could get some validation/invalidation.

If $\displaystyle X$ is connected Hausdorff and $\displaystyle \varphi:X\mapsto Y$ continuous then is $\displaystyle \Gamma_\varphi\subseteq X\times Y$ connected? ($\displaystyle \Gamma_\varphi$ is the graph of $\displaystyle \varphi$). I think the answer is yes.

Proof: We first need a lemma

Lemma: Let $\displaystyle X$ be connected Hausdorff, then $\displaystyle \Delta_X\subseteq X\times X$ is connected ($\displaystyle \Delta_X$ is the diagonal).
Proof: Clearly the map $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ given by $\displaystyle x\mapsto (x,x)$ is continuous. But, it is not hard to see that $\displaystyle \Delta_X=\left(\iota_X\oplus \iota_X\right)\left(X\right)$. But, connectedness is invariant under continuous mapping. So, the conclusion follows $\displaystyle \blacksquare$

Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since $\displaystyle \varphi$ and the identity map $\displaystyle \iota_X$ are continuous we have that $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ is continuous. I now claim that $\displaystyle \Gamma_\varphi=\left(\iota_X\times\varphi\right)\l eft(\Delta_X\right)$.

To see this, let $\displaystyle (y,y')\in\Gamma_\varphi$ then $\displaystyle (y,y')=(x,\varphi(x))$ for some $\displaystyle x\in X$. Clearly then $\displaystyle \left(\iota_X\times\varphi\right)\left((x,x)\right )=\left(x,\varphi(x)\right)=(y,y')$ and so $\displaystyle (y,y')\in \left(\iota_X\times\varphi\right)\left(\Delta_x\ri ght)$

Conversely, let $\displaystyle (y,y')\in\left(\varphi\times\iota_X\right)\left(\D elta_X\right)$. Then, $\displaystyle (y,y')=\left(\varphi\times \iota_X\right)\left((x,x)\right)$ for some $\displaystyle x\in X$ but that means that $\displaystyle (y,y')=(x,\varphi(x))$ and so $\displaystyle (y,y')\in\Gamma_{\varphi}$.

Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. $\displaystyle \blacksquare$

Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.
• Mar 11th 2010, 05:54 PM
Drexel28
Idiot moment. Of course it's true.

$\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ and $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ are both continuous. So, of course any topological property invariant under continuous mappings transfers from $\displaystyle X$ to $\displaystyle \Delta_X$ and $\displaystyle \Gamma_\varphi$

Sorry for the hiccup :)
• Mar 11th 2010, 06:31 PM
Jose27
Quote:

Originally Posted by Drexel28
I came up with the following question, and would appreciate if I could get some validation/invalidation.

If $\displaystyle X$ is connected Hausdorff and $\displaystyle \varphi:X\mapsto Y$ continuous then is $\displaystyle \Gamma_\varphi\subseteq X\times Y$ connected? ($\displaystyle \Gamma_\varphi$ is the graph of $\displaystyle \varphi$). I think the answer is yes.

Proof: We first need a lemma

Lemma: Let $\displaystyle X$ be connected Hausdorff, then $\displaystyle \Delta_X\subseteq X\times X$ is connected ($\displaystyle \Delta_X$ is the diagonal).
Proof: Clearly the map $\displaystyle \iota_X\oplus\iota_X:X\mapsto X\times X$ given by $\displaystyle x\mapsto (x,x)$ is continuous. But, it is not hard to see that $\displaystyle \Delta_X=\left(\iota_X\oplus \iota_X\right)\left(X\right)$. But, connectedness is invariant under continuous mapping. So, the conclusion follows $\displaystyle \blacksquare$

Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since $\displaystyle \varphi$ and the identity map $\displaystyle \iota_X$ are continuous we have that $\displaystyle \iota_X\times\varphi:X\times X\mapsto X\times Y$ is continuous. I now claim that $\displaystyle \Gamma_\varphi=\left(\iota_X\times\varphi\right)\l eft(\Delta_X\right)$.

To see this, let $\displaystyle (y,y')\in\Gamma_\varphi$ then $\displaystyle (y,y')=(x,\varphi(x))$ for some $\displaystyle x\in X$. Clearly then $\displaystyle \left(\iota_X\times\varphi\right)\left((x,x)\right )=\left(x,\varphi(x)\right)=(y,y')$ and so $\displaystyle (y,y')\in \left(\iota_X\times\varphi\right)\left(\Delta_x\ri ght)$

Conversely, let $\displaystyle (y,y')\in\left(\varphi\times\iota_X\right)\left(\D elta_X\right)$. Then, $\displaystyle (y,y')=\left(\varphi\times \iota_X\right)\left((x,x)\right)$ for some $\displaystyle x\in X$ but that means that $\displaystyle (y,y')=(x,\varphi(x))$ and so $\displaystyle (y,y')\in\Gamma_{\varphi}$.

Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument. $\displaystyle \blacksquare$

Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.

Please, maps to is not equal to a function...please...you're driving me crazy...

• Mar 11th 2010, 06:34 PM
Drexel28
Quote:

Originally Posted by Jose27
Please, maps to is not equal to a function...please...you're driving me crazy...

Take that! :)
• Mar 11th 2010, 10:44 PM
Jose27
Quote:

Originally Posted by Drexel28

Okay, I guess it's my bad since I didn't specify what I was ranting about...(sorry it came off a bit insulting)

What I'm talking about is that usually one writes a function with $\displaystyle \rightarrow$ and the correspondence relation with $\displaystyle \mapsto$ (not trying to impose, but it really bugs me out...I really need to relax a bit with notation...(Speechless))
• Mar 12th 2010, 10:51 AM
Drexel28
Quote:

Originally Posted by Jose27
Okay, I guess it's my bad since I didn't specify what I was ranting about...(sorry it came off a bit insulting)

What I'm talking about is that usually one writes a function with $\displaystyle \rightarrow$ and the correspondence relation with $\displaystyle \mapsto$ (not trying to impose, but it really bugs me out...I really need to relax a bit with notation...(Speechless))

It's ok. I appreciate that you'd let met know. :)

I understand that one usually puts $\displaystyle \phi:G\longrightarrow G'$ and $\displaystyle g\mapsto g'$. But in my naieve ignorance I got used to doing it the wrong way. It's just kind of persisted. I'll eventually wise up