Some introductory Topology questions

[1MileCrash]

Hi all,

My Topology textbook arrived in the mail today, so I started reading it. It begins with an introduction to an object called metric spaces.

It says

A metric on a set X is a function d: X x X -> R that satisfies the following conditions:

-some conditions--

I am not completely sure about this notation (mainly the "d:" part.) I believe that X x X is simply the Cartesian product. My guess is that d is the function's name, and the rest just says the domain of this function in relation to the set X (kinda?)

So is it correct to read it in this way: d: X x X -> R

Means:

Some function d, of x and y, where x and y refer to axis of a 2d plane. This 2d plane is the cartesian product of the set X and itself.

So if X were the set {1,2,3}, this notation defines d to be a function of x and y such that either x or y is equal to 1, 2, or 3?

Or is this not right?

 

[mfb]

That is right.

With X={1,2,3}, you could consider d(1,1), d(1,2), d(1,3), d(2,1), ..., d(3,3) as function values, for example.

 

[1MileCrash]

Cool, thank you.

This will be fun :)

 

[micromass]

Some function d, of x and y, where x and y refer to axis of a 2d plane. This 2d plane is the cartesian product of the set X and itself.

Not sure where you get a 2d plane from. If ##X=\mathbb{R}##, then ##X\times X = \mathbb{R}^2## and that's certainly a 2d plane. But ##X## can be something far more general. For example, ##X=\{\text{Obama}, \text{Christmas Tree}\}##. I don't see how ##X\times X## is some kind of plane then.

 

[Office_Shredder]

I am not completely sure about this notation (mainly the "d:" part.) I believe that X x X is simply the Cartesian product. My guess is that d is the function's name, and the rest just says the domain of this function in relation to the set X (kinda?)

It says the domain and the codomain of the function (the R being the codomain). The general notation is
[tex] f: A \to B [/tex]
is a function whose domain is A and whose codomain is B (which means it always takes a value in B - this doesn't mean every point in B can be reached!). Examples:
[tex] \arctan(x): \mathbb{R} \to \mathbb{R} [/tex]
[tex] \arctan(x): \mathbb{R} \to (-\pi/2,\pi/2) [/tex]
[tex] \sqrt{x}: [0,\infty) \to \mathbb{R} [/tex]
[tex] \sqrt{x}: [0,\infty) \to [0,\infty)[/tex]
[tex] \cos(x): (-\infty,\infty) \to [-1,1] [/tex]
[tex] \sin(x): (-\pi/2,\pi/2) \to (-1,1)[/tex]

Non-examples:
[tex] \sqrt{x}: \mathbb{R} \to \mathbb{R} [/tex]
[tex] \arctan(x): \mathbb{R} \to (-1,1) [/tex]

 

[1MileCrash]

It says the domain and the codomain of the function (the R being the codomain). The general notation is
[tex] f: A \to B [/tex]
is a function whose domain is A and whose codomain is B (which means it always takes a value in B - this doesn't mean every point in B can be reached!).

So..

What that says is that A is the domain, and that B is some type of "category" that the elements of A belong to? (A is the set of all possible arguments of the function, and it is a subset of B, and B is some category to which all possible arguments belong too (like the real numbers?))

 

[1MileCrash]

Not sure where you get a 2d plane from. If ##X=\mathbb{R}##, then ##X\times X = \mathbb{R}^2## and that's certainly a 2d plane. But ##X## can be something far more general. For example, ##X=\{\text{Obama}, \text{Christmas Tree}\}##. I don't see how ##X\times X## is some kind of plane then.

I would still think of that as 2 dimensional. If the use of the word "plane" requires both axes to be continuous then I guess I would call that a 2d "grid" and not plane. Is that what you mean?

 

[micromass]

I would still think of that as 2 dimensional. If the use of the word "plane" requires both axes to be continuous then I guess I would call that a 2d "grid" and not plane. Is that what you mean?

The thing is that "two dimensional" and "plane" refer to very specific situations. You shouldn't use those words in other situations. I know what you mean though.

 

[micromass]

So..

What that says is that A is the domain, and that B is some type of "category" that the elements of A belong to? (A is the set of all possible arguments of the function, and it is a subset of B, and B is some category to which all possible arguments belong too (like the real numbers?))

Haven't you learned about functions or the ##f:A\rightarrow B## notation before? If so, it's something you need to learn before starting topology.

 

[1MileCrash]

Haven't you learned about functions or the ##f:A\rightarrow B## notation before? If so, it's something you need to learn before starting topology.

I did but a long time ago in an introductory proof class.

EDIT: I think my question was riddled with confusion. B should be related to the range of the function right? So the set of all possible outputs of the function is a subset of the codomain?

 

[HallsofIvy]

Yes, the range is a subset of the co-domain.

 

[1MileCrash]

Ok, one more question. Is the codomain noted because the actual range can sometimes be an ugly looking set, and providing the codomain in the notation tells us enough about this function?

 

[Office_Shredder]

Essentially. The codomain is just telling you what kind of objects the function gives you - for example if I just told you a metric is a function defined on [itex] X \times X[/itex], and it has the following property: [itex] d(x,y) \geq 0 [/itex] the first thing you should ask me is "what the heck does [itex] \geq [/itex] mean? What set is d(x,y) contained in and what is the ordering and why is there a zero?"

Questions which are moot if I tell you the codomain is [itex] \mathbb{R}[/itex].

The reason why a codomain is usually not the same as the range when the function is defined is because it's not obvious what the range is when the function is defined, but a codomain is required for the function to have any meaning (you have to specify what the output looks like). Notice that the codomain is fairly arbitrary - I could say the codomain of the square root function is R, or the non-negative real numbers, or the interval from -3 to infinity and those would all be acceptable

 

[1MileCrash]

Understood, thanks all! May be back with more questions as I keep going. :)

 

[1MileCrash]

Another question!

My book mentions that d is a "distance function." Does this mean that the codomain of d will always be the reals (I know of no other way to express a "distance") for any metric/metric space?

 

[micromass]

Another question!

My book mentions that d is a "distance function." Does this mean that the codomain of d will always be the reals (I know of no other way to express a "distance") for any metric/metric space?

By definition, a distance function or metric is a function ##d:X\times X\rightarrow \mathbb{R}^+##. So the codomain are the nonnegative reals.

 

[1MileCrash]

Alright. My book just puts R, but includes a positive or zero result as a property of a metric, so I guess those are equivalent.

 

[1MileCrash]

Also, I think I understand what you were saying before regarding the 2d plane.

XxX -> R

Means it goes from the set XxX, which just so happens to be a set of ordered pairs, to the set of real numbers. I don't need to involve the idea of planes, really.

 

[1MileCrash]

I finally got to the exercises. The first one is proving that a certain d is a metric on set X. It trivially is, based on what I see, but I want to make sure that I wrote a proof that is reasonably coherent.

Also, where I wrote "This is only possible if... = 0, since d is non-negative" was kind of stupid. It's really because 0 is the only other possible value of d.

If someone that can read my handwriting could just take a look.. I would appreciate it!

EDIT: I see that the pic got resized, try:

http://postimg.org/image/qqelnlcrb/

 

[mfb]

I would try to use less words, but the ideas are right.

 

[1MileCrash]

I just did two more proofs, proving that the taxicab metric was a metric, and proving the minkowski inequality with Cauchy schwarts inequality.

The first was pretty easy, but it led me to a question I will post about later.

The second took a few hours (I know, I suck) but it was one of the most badass proofs I've ever done when I finally saw the result unfold. I'll have to rewrite it because there are a lot of scratch outs and stuff, but I will post them!

 

[micromass]

Proving Minkowski is actually not so easy, so don't worry if it took you a while. It's indeed quite elegant. Good job that you managed to prove it!

 

[1MileCrash]

Thanks, it was quite cool. My trouble was getting a dot product to appear ( and realizing it was there once I did ).

Here's what I did to prove it, among the scratch work.

http://oi42.tinypic.com/2hey6v9.jpg

 

[micromass]

Pretty nice proof. Very impressive that you found it yourself!

It's not the standard proof though, here is the proof that is usually given for this:

[tex]
\begin{eqnarray*}
\|\mathbf{x} + \mathbf{y}\|_2^2
& = & \sum_{i=1}^n (x_i + y_i)^2\\
& = & \sum_{i=1}^n x_i(x_i + y_i) + \sum_{i=1}^n y_i(x_i + y_i)\\
& = & \mathbf{x} \cdot (\mathbf{x} + \mathbf{y}) + \mathbf{y}\cdot (\mathbf{x} + \mathbf{y})\\
& \leq & \|\mathbf{x}\|_2 \|\mathbf{x} + \mathbf{y}\|_2 + \|\mathbf{y}\|_2 \|\mathbf{x} + \mathbf{y}\|_2\\
& = & (\|\mathbf{x}\|_2 + \|\mathbf{y}\|_2) \|\mathbf{x} + \mathbf{y}\|_2
\end{eqnarray*}
[/tex]

Now divide both sides by ##\|\mathbf{x} + \mathbf{y}\|_2## and you obtain the inequality.

 

[Fredrik]

That last picture takes 40 seconds to load. I didn't see any of the first ones, because I gave up much earlier than that. So I have to suggest that when you have to use an image hosting site, use another one.

The option that micromass is using is of course much better than using images: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

 

[1MileCrash]

Pretty nice proof. Very impressive that you found it yourself!

It's not the standard proof though, here is the proof that is usually given for this:

[tex]
\begin{eqnarray*}
\|\mathbf{x} + \mathbf{y}\|_2^2
& = & \sum_{i=1}^n (x_i + y_i)^2\\
& = & \sum_{i=1}^n x_i(x_i + y_i) + \sum_{i=1}^n y_i(x_i + y_i)\\
& = & \mathbf{x} \cdot (\mathbf{x} + \mathbf{y}) + \mathbf{y}\cdot (\mathbf{x} + \mathbf{y})\\
& \leq & \|\mathbf{x}\|_2 \|\mathbf{x} + \mathbf{y}\|_2 + \|\mathbf{y}\|_2 \|\mathbf{x} + \mathbf{y}\|_2\\
& = & (\|\mathbf{x}\|_2 + \|\mathbf{y}\|_2) \|\mathbf{x} + \mathbf{y}\|_2
\end{eqnarray*}
[/tex]

Now divide both sides by ##\|\mathbf{x} + \mathbf{y}\|_2## and you obtain the inequality.

That's very clear. I don't think I would have thought to do that on my own, though.

Why do you put a subscript by your norms?

 

[1MileCrash]

That last picture takes 40 seconds to load. I didn't see any of the first ones, because I gave up much earlier than that. So I have to suggest that when you have to use an image hosting site, use another one.

The option that micromass is using is of course much better than using images: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Sorry, I'm on a mobile at the moment and had to commandeer my dad's cellphone to read micromass's markup. When I get back home I can type it up proper.

 

[Fredrik]

When I get back home I can type it up proper.

This is not necessary if your questions have been answered. Just keep it in mind for the next time you ask a question.

 

[1MileCrash]

Another question (that doesn't require latex)

I am trying to prove that the Euclidean metric is a metric.

It is defined as the root of a number that is always positive (or zero).

And to prove it is a metric, it has to be nonnegative.

How do I handle the root? I assumed I don't just say "take the positive root so that my proof works."

Or does the definition of the euclidean metric already imply that we are taking the positive root?

 

[Fredrik]

The square root of a non-negative real number y is defined as the non-negative real number x such that ##x^2=y##.

 

[1MileCrash]

I'm having trouble googling this one..

What is |R^(omega)?

 

[micromass]

I'm having trouble googling this one..

What is |R^(omega)?

Which book are you doing? Munkres?

 

[1MileCrash]

No, it's my textbook for the class. "Foundations of Topology" C. Wayne Patty

 

[micromass]

No, it's my textbook for the class. "Foundations of Topology" C. Wayne Patty

Never heard of that. Well, ##\mathbb{R}^\omega## is usually defined as the set of sequences in ##\mathbb{R}##. So

[tex]\mathbb{R}^\omega = \{(x_n)_n~\vert~x_n\in \mathbb{R}\}[/tex]

This is the definition Munkres uses. I think your book uses the same definition.

 

[1MileCrash]

My book gives a definition in an appendix that seems to match what you're saying.

Unfortunately, I'm having a hard time grasping the idea (the set you've wrote, I can read, but I can't "see it."). It is the set of all possible sequences of any length in R? Is that like a power set of R, but where order matters for the elements? Or am I way off?