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[Kidphysics]

This last video I just watched makes me wonder if I am missing something or
perhaps he just made a rather large error. Someone else in the comments pointed out the mistake as well. Was there a mistake in my understanding or was there a mistake in the proof?



thanks in advance guys

 

[henry_m]

I think the lecturer rather unfortunately swapped the names of his metrics about 8 minutes in, which is very confusing! But the gist of the proof remains correct. The point is that if you draw a circle (ball in Euclidean metric) with radius sqrt(2)r, the square of side 2r (ball in max metric) will fit inside it. This should be clear from drawing a picture.

 

[Kidphysics]

I think the lecturer rather unfortunately swapped the names of his metrics about 8 minutes in, which is very confusing! But the gist of the proof remains correct. The point is that if you draw a circle (ball in Euclidean metric) with radius sqrt(2)r, the square of side 2r (ball in max metric) will fit inside it. This should be clear from drawing a picture.

Please allow me to understand the steps that were taken. If i am understanding correctly...

|x1-y1|^2+|x2-y2|^2= d^2(x,y) where d= euclidean metric (that we squared)

now the max metric takes the maximum of either one of these. However, then the max metric would only have one term, where the euclidean has two so we multiply the greater term by two. Now the max metric is still greater than the euclidean metric! Perhaps if we multiplied the Euclidean metric by sqrt(2) and left the max metric radius at 1 then they would become equal but to multiply something that is already larger by two and something that is already smaller by sqrt(2) does not help me understand what I know we want (the Euclidean ball inside the max ball)...

 

[henry_m]

Okay, so let's call the Euclidean metric $d_E$, and the max metric $d_M$ to avoid confusion. We have straightforwardly that for all x,y, $d_M(x,y)\leq d_E(x,y)$, which is the statement that the Euclidean ball fits inside the 'max' ball. We want to also show that $d_E(x,y)\leq \sqrt{2} d_M(x,y)$. This is the statement that the 'max' ball fits inside the (bigger) Euclidean ball.

It's equivalent to prove the square of this, since everything is positive: $d_E(x,y)^2\leq 2 d_M(x,y)^2$. WLOG, we can assume $|x_1-y_1|\geq|x_2-y_2|$. Then $d_E(x,y)^2=(x_1-y_1)^2+(x_2-y_2)^2\leq 2 (x_1-y_1)^2=2d_M(x,y)^2$.

I agree that the video was confusing here. The best way to see what's going on is to draw the picture and let that guide you through the proof.

 

[Kidphysics]

We have straightforwardly that for all x,y, $d_M(x,y)\leq d_E(x,y)$, which is the statement that the Euclidean ball fits inside the 'max' ball.

this may be the root of my confusion. I know that the Euclidean ball fits inside of the max ball.. but I was under the assumption that $d_M(x,y)\geq d_E(x,y)$ since the ball constructed by the max metric is larger than the ball constructed by the Euclidean metric. The reason the circle fits inside the square (I have just started in topology).

I understand the proof and how the max metric 'steals' the largest distance and duplicates it (which is basically what the max metric does, takes the largest distance) to insure it being greater and at worst equal to the Euclidean ball. But wasn't the Max metric always greater? Wasn't the Euclidean ball always in the Max ball? Wouldn't we have to multiply the radius by some constant to the Euclidean radius so that now the square is in the circle?

[PLAIN]http://img51.imageshack.us/img51/6023/problemb.png

as you can tell I want to understand this

 

[henry_m]

this may be the root of my confusion. I know that the Euclidean ball fits inside of the max ball.. but I was under the assumption that $d_M(x,y)\geq d_E(x,y)$ since the ball constructed by the max metric is larger than the ball constructed by the Euclidean metric.

I see your confusion. The inequality is the other way round, which is perhaps a little counterintuitive. Think of it like this:

We want to show that the Euclidean ball at x fits inside the max ball at x. So we want to show that every y in the Euclidean ball is also in the max ball. So that means we want to show that if $d_E(x,y)<r$ we also have $d_M(x,y)<r$. But $d_M(x,y)\leq d_E(x,y)<r$ so we're done.

To see why $d_M(x,y)\leq d_E(x,y)$, look at their squares. The squared Euclidean metric has two terms, and the square of the max metric is equal to the bigger of those two terms. But that still leaves one term over, so the Euclidean distance is the larger one.

Think about the points which are in the max ball (the square) but not in the Euclidean ball (the circle). That means their Euclidean distance from the centre must be larger than r, though there max distance is less than r. This is another clue that the inequality is the right way round.

 

[Kidphysics]

To see why $d_M(x,y)\leq d_E(x,y)$, look at their squares. The squared Euclidean metric has two terms, and the square of the max metric is equal to the bigger of those two terms. But that still leaves one term over, so the Euclidean distance is the larger one.

This indeed does make sense.

Think about the points which are in the max ball (the square) but not in the Euclidean ball (the circle). That means their Euclidean distance from the centre must be larger than r, though there max distance is less than r. This is another clue that the inequality is the right way round.

I think what is most confusing is the fact that we are plotting the max metric on a Euclidean plane. Looking at the definitions of the two metrics (and squaring) I can see how the Euclidean metric would have a greater radius.. but when plotting on a graph it seems like since the plot of the max metric only takes into account left right up and down, it seems to have a larger set of points! If we draw out a random set of points and draw the balls given by the value of the different radii then the biggest ball will be the one with the bigger radius- the Euclidean ball. Why is it that when the Max metric get's plotted it seems to have more points than the Euclidean ball, yet it is supposed to be smaller and contain an equal if not less amount?

Thank you for all your help thus far

 

[henry_m]

If one measure of distance is larger than another for any two given points, then the balls it describes must be smaller, not larger. Try this analogy:

Let our two metrics be the 'walking' metric and the 'cycling' metric, which they tell us the time in minutes to get between two points by foot or by bike. The bike is faster, so the walking metric between any two points gives us a larger measure of distance. And given an hour, we can get much further on a bike: the balls for the cycling metric are larger.

 

[Kidphysics]

If one measure of distance is larger than another for any two given points, then the balls it describes must be smaller, not larger. Try this analogy:

Let our two metrics be the 'walking' metric and the 'cycling' metric, which they tell us the time in minutes to get between two points by foot or by bike. The bike is faster, so the walking metric between any two points gives us a larger measure of distance. And given an hour, we can get much further on a bike: the balls for the cycling metric are larger.

Okay this is very interesting to me. A problem I'm having is that when I see the box (max metric) I see that there are some points in it that are not contained by the circle. In set theory I think of this as a greater cardinality of ordered pairs? In the set of all points contained by the max metric set.. This could also is a reason that I have been equating larger radius to larger distance.. Although the idea that the cycling metric is bigger because the "range" of the metric per given time "unit"