1. Pathwise connected

My professor has the idea of a subset being pathwise connected in his notes but he never talked about it. I would appreciate any help.

He says...
Let $D$ be a subset of complex numbers. $D$ is pathwise connected if given any two points $a,b\in D$, there exists a continuous function $\varphi :[0,1]\rightarrow D$, such that $\varphi (0)=a$ and $\varphi (1)=b$.

He then asks us to prove that a rectangle is piecewise connected.

2. You might want to have a look at Wikipedia for a little illustration: Connected space - Wikipedia, the free encyclopedia

The idea is simple: your set is path-connected if for any two points in the set, a and b, you can join them with a curve.

For your problem, take two points from the rectangle and come up with a curve joining them. (Don't think hard about it - there is a very straightforward curve.)

3. Can't a straight line connect any two points in a rectangle? So I just have to make a line that connects these two points and then the rectangle is pathwise connected?
So would a counter example be a function f that is on the interval [0,1]U[2,3] since a curve cannot connect all points and be continuous?

4. Originally Posted by zebra2147
Can't a straight line connect any two points in a rectangle? So I just have to make a line that connects these two points and then the rectangle is pathwise connected?
So would a counter example be a function f that is on the interval [0,1]U[2,3] since a curve cannot connect all points and be continuous?
To your counter example that set isn't connected and thus can't be path connected. The Topologist' Sine Curve
is an example of a connected but not path connected space.

For the first part what if you took $\displaystyle [a,b]\times [c,d]=\bigcup_{x\in[a,b]}\left(\{x\}\times[c,d]\cup[a,b]\times\{c\}\right)$

5. ok, that makes sense. So if I have to prove that any pathwise connected subset of R is an interval...
This should be relatively simple. I just have to show that since the subset is pathwise connected, then it is also connected. Therefore, all the connected points lie in an interval???

6. Originally Posted by zebra2147
ok, that makes sense. So if I have to prove that any pathwise connected subset of R is an interval...
This should be relatively simple. I just have to show that since the subset is pathwise connected, then it is also connected. Therefore, all the connected points lie in an interval???
Yes. Recall that a topological space $X$ is not connected if and only if there exists a continuous surjection $f:X\to \{0,1\}_D$ where $\{0,1\}_D$ is the two points discrete space. So, suppose that $Y$ is path connected but not connected. Then, there exists some continuous surjection $f:Y\to\{0,1\}_D$. So, let $x\in f^{-1}(\{0\}),y\in f^{-1}(\{1\})$. Then, by assumption there exists some path $p:[0,1]\to Y$ which connects them (i.e. $p(0)=x$ and $p(1)=y$) Then, the map $p\circ f:[0,1]\to \{0,1\}_D$ is continuous (being the composition of continuous maps) and surjective since $p(f(0))=p(x)=0$ and $p(f(1))=p(y)=1$. But...what's wrong with this picture?

7. ummm... I'm guessing there might be a contadiction somewhere???
I don't really see it though.
Maybe that we have $p(x)=0$and $p(0)=x$?? I'm pretty sure that is not even close to right though.

8. Originally Posted by zebra2147
ummm... I'm guessing there might be a contadiction somewhere???
I don't really see it though.
Maybe that we have $p(x)=0$and $p(0)=x$?? I'm pretty sure that is not even close to right though.
I stated earlier that a space is disconnected if and only if there exists a continuous surjection onto the two point discrete space. Read the last sentence in my last post again with that in mind.

9. Well in the last line you state that $p\circ f:[0,1]\to \{0,1\}_D$ is continuous and surjective. Thus, it must be disconnected.

10. Originally Posted by zebra2147
Well in the last line you state that $p\circ f:[0,1]\to \{0,1\}_D$ is continuous and surjective. Thus, it must be disconnected.
Thus, $[0,1]$ must be disconnected...but....that's just plain stupid, right?

P.S. If this whole continuous surjections onto the two-point discrete space stuff is confusing let me know what you use for connectedness.

11. Right. We know that [0,1] shouldn't be disconnected. And I think that the reason I probably seem oblivious to many of your conclusions is because my professor hasn't used the terms "two-point discrete" or "connectedness". I still get the general idea though and your help is very much appreciated.

12. Not to revisit an old topic but I was looking over the proof that a rectangle is piecewise connected. Drexel28 recommended looking at $\displaystyle [a,b]\times [c,d]=\bigcup_{x\in[a,b]}\left(\{x\}\times[c,d]\cup[a,b]\times\{c\}\right).$ At the time it made sense but now that i look at it I'm kinda stuck. I'd appreciate any further guidance.

13. Originally Posted by zebra2147
Not to revisit an old topic but I was looking over the proof that a rectangle is piecewise connected. Drexel28 recommended looking at $\displaystyle [a,b]\times [c,d]=\bigcup_{x\in[a,b]}\left(\{x\}\times[c,d]\cup[a,b]\times\{c\}\right).$ At the time it made sense but now that i look at it I'm kinda stuck. I'd appreciate any further guidance.
So, (I hope I wrote the union right) let me give you an analogous problem and see if you can relate the proofs

So, I want to prove that the closed unit disk $\overline{\mathbb{D}}$ is path connected. Well, there are two obvious ways to do it A) it's convex and B) what we care about.

So, we know that in general if $\left\{A_{\alpha}\right\}_{\alpha\in\mathcal{A}$ is a class of path connected subspaces of $X$ and $\displaystyle \bigcap_{\alpha\in\mathcal{A}}A_{\alpha}\ne\varnot hing$ then $\displaystyle \bigcup_{\alpha\in\mathcal{A}}A_{\alpha}$ is connected. So, if we can write $\overline{\mathbb{D}}$ as the union of intersecting path connected sets it must be path connected. So, the obvious choice would be to take the union of all circles...but unfortuantely they're disjoint. So, we add in something extra to each that "connects" them.

So, clearly the line $[0,1]\times\{0\}$ is path connected since $[0,1]\times\{0\}\approx[0,1]$. Moreover, each circle centered at the origin $S_r(0)=\left\{z\in\mathbb{C}:\|z\|=r\right\}$ is path connected since $S_r(0)\approx\mathbb{S}^1$ (the unit circle) and $\mathbb{S}^1$ is path connected since it's the image of $e^{2\pi i t}$ under $[0,1]$. Thus, since $S_r(0)\cap\left( [0,1]\times \{0\}\right)\ne\varnothing$ we can conclude that $S_r(0)\cup\left([0,1]\times\{0\}\right)$ is path connected. Note then that $\displaystyle \overline{\mathbb{D}}=\bigcup_{0 and since $\displaystyle [0,1]\times\{0\}\subseteq \bigcap_{0 it follows that $\overline{\mathbb{D}}$ can be written as the union of intersecting path connected subsets of $\mathbb{C}$ and is thus path connected.

Now, forget all the extra fluff. We wanted to write our space as the union of intersecting path connected spaces, and so we took the obvious example and stuck in an extra "connection" (the radial line). Clearly the rectangle can be thought of as the union of all the "vertical lines" contained in it, and these are all path connected. But, they're disjoint. So, try adding in (or seeing what I added in) to make them not disjiont.

14. Alright so trying to mimic your example...
We want to show that $\{x\}\times[c,d]$ and $[a,b]\times\{c\}\right)$ are both path connected and that $\bigcap_{x\in[a,b]}\left(\{x\}\times[c,d]\cup[a,b]\times\{c\}\right)\not= \emptyset$.

I know I didn't show they are not disjoint... I'll post as soon as I have a better idea.

15. Originally Posted by zebra2147
Alright so trying to mimic your example...
We want to show that $\{x\}\times[c,d]$ and $[a,b]\times\{c\}\right)$ are both path connected and that $\bigcap_{x\in[a,b]}\left(\{x\}\times[c,d]\cup[a,b]\times\{c\}\right)\not= \emptyset$.

I know I didn't show they are not disjoint... I'll post as soon as I have a better idea.
Ok, good luck. Two suggestions though

a) In general $\{x\}\times X\approx X$

b) Since $c$ is fixed $[a,b]\times\{c\}$ is "constant" throughout the union.

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