$$\text { let } D\subset \mathbb R, f:D\rightarrow R \text { be continuous} $$ prove D is connected if
$$\{(x,f(x)):x\in D\}$$ is connected in
$$R^2$$

Question

$$\text { let } D\subset \mathbb R, f:D\rightarrow R \text {  be continuous}  $$ prove D is connected if 
$$\{(x,f(x)):x\in D\}$$ is connected in 
$$R^2$$

jamesj · Answer

What definition of connected do you have?

anonymous · Answer

might as  well say D is an interval since it is in R right?

jamesj · Answer

Yes, but that won't work in R^2.  So what definition are you working with?

anonymous · Answer

oh i mean i need D is an interval, not the graph right

anonymous · Answer

not disconnected,
not the union of two disjoint open sets
contains no non trivial clopen sets

jamesj · Answer

ok, good.

anonymous · Answer

haven't looked at this stuff in 20+ years, so any hint would be appreciated

jamesj · Answer

Thinking ...

anonymous · Answer

by all means take your time. i am not sure if even what to use.  mean value?

jamesj · Answer

No, no.  Definitely not.  You want to use the topological definition of continuity here; or at least I'm hypothesizing we do.  I'm looking for the 1-2 line proof that uses it.

anonymous · Answer

that would be nice.

jamesj · Answer

Are you sure you've got the if going in the right direction.  That is, let's call the set G (for graph).  Are you asked to show that 
G connected => D connected
or 
D connected => G connected

anonymous · Answer

problem is copied verbatim i think let me check again

anonymous · Answer

yes it says prove D is  connected IF the graph is a connected subset of R^2

anonymous · Answer

so first one

jamesj · Answer

ok.   Define a function p : G --> D by p(x,y) = p(x).  p is projection and it is clearly onto D.  Further, p is continuous.  This last statement isn't completely obvious, but it is true given the definition of G and the fact that f is continuous.  This is a critical step and I leave that as a lemma for you to show using epsilon-delta formality.

Given that, let's flip now to the topological definition of continuity (which is equivalent to the e-d definition for real functions, hopefully a result you already have.)

Suppose that D was not connected and $ D = A \cup B $ was a separation of D into two disjoint non-empty sets open in D.  Then $ p^{-1}(A) $ and $ p^{-1}(B) $ are disjoint sets; they are open in G because p is continuous, and non-empty because g is surjective.  Therefore they form a separation of G, contradicting the assumption that G is connected. qed.

jamesj · Answer

"...non-empty because p (not g) is surjective ..."

anonymous · Answer

wow thanks. so last step is to show that the projection is continuous.   is continuous  from D to R, the graph of f is connected implies D is connected.
and inverse image of open set is open is definition of continuity, so fine with that.  thank, i am not sure i would have come up with this and certainly not in 20 minutes!

anonymous · Answer

that came out garbled for some reason, i mean only this part 
wow thanks. so last step is to show that the projection is continuous.  

and inverse image of open set is open is definition of continuity, so fine with that.  thank, i am not sure i would have come up with this and certainly not in 20 minutes!

jamesj · Answer

Yes.  I knew the continuous image of a connected set was connected; I just had to figure out how to use that result here. 

I'm off to bed now but I'll check in tomorrow in case you get stuck.  

Nice problem.

anonymous · Answer

thanks again,  i am off to bed as well (too old to stay up so late!)

jamesj · Answer

There's an easy proof p is continuous.  

The projection $ p : \mathbb{R}^2 \to \mathbb{R} $ defined by $ p(x,y) = x  $ is continuous (why?) and then the restriction of p to G, $ p|_G $, is therefore continuous where G takes the subset topology of $ \mathbb{R}^2 $.

The fact we can write this underlines something rather strange about the result: IF G is connected then D is connected.  In other words, we haven't used continuity of f at all.

Ok then.  But when is G connected?  When f is continuous.  This is the more or less the intermediate value theorem.  But the question and all of our results to date haven't told us that.

So I'm inclined to think of this question as an exercise in topology, but not a very useful result.

If you have a tutorial where you have a change to discuss this question, I'd be interested to hear what your tutor has to say.