I love your videos. One question, for g to be a metric shouldn't it also have a 3rd property, satisfying the triangular inequality? g(x,y) + g(y,z) >= g(x,z)
Yes I forgot to mention this property! I was going to address this when I make a video about the metric tensor however, but thanks for mentioning it here!
Great series of videos! A couple of questions: 1.) The metric operates on a pair of vectors, correct? But we haven't endowed our sets here with a vector property as far as I can tell. Or is that implicit in defining a topological space? 2.) Does the metric always have to map to the reals? Or can more exotic metrics be constructed that map to e.g. C or N? 3.) Won't the Hausdorff property always be true unless you somehow include two copies of a point in your set? Struggling to understand the intuition behind this definition.
1) So I may have gotten ahead of myself here and begun referring to the metric tensor by mistake, which does act on a pair of vectors to give a notion of `distance', however already relying on additional structure that we don't yet have (vectors). The metric in this case simply refers to a function of pairs of points in the space. 2) No, more exotic metrics could easily be considered but are not particularly useful for our case working over R, indeed, the case of a discrete topology would have to have a metric in N! 3)We usually would like Hausdorff to be true, at least for everything that would be considered `standard' and for everything else following (manifolds). The intuition is pretty obscured unless you really dig deep into discussing limits and continuity, so for our purposes we can just regard it as a `separation axiom' that effectively lets us talk about the points of a topological space as being `unique', in the sense that they are all surrounded by their own open balls that do not intersect. Sorry I've taken so long to reply, hope that is somewhat helpful feel free to ask more!
In principle I don't see why not, although I'm not familiar with any cases where this is used, since we need to have the positive definite condition, allowing for complex values can complicate this considerably.
i finish this lecture. i get confused (but not very ) when you represents the open ball with x i and y i (in lecture 3 ) since there is a open ball for only one centre if you represents your xi in R^2 (lecture #3) or R ,why do you write something like : B(xi) = { yi : sqrt ( sum i ( xi -yi )^2) < r } i think xi should be only c or xj becuase if you sum over the yi- xi then you are not summing only one xi only ? ( not a lot of yi around one xi at the centre but rather a pair of yi and xi ) i think the "sum" terms can be omitted since there is no point to sum up all yi in the argument for all yi such that : |yi(an element in R^2) - x (an centre element in R^2)|
Yes you are right it is alot easier to just use the modulus definition to define the ball. Maybe I was unclear in how I used the sum symbol however and should have used separate indices, the indices here are only referring to the particular coordinate in R^n. It might have been better to just call the centre of the circle c rather than x_i, but I gave it an index so that it's clear each component of the centre is subtracted from the correct component of the y_i coordinate If the sum formula is unclear just consider the case for R^2 and let x_i={0,0}. If our coordinates for the points in the ball are y_i=(y1,y2) then B(0,0) = sqrt[sum(y_i -x_i)^2] = sqrt[(y1 - 0)^2+(y2 - 0)^2] < r Which is the circle of radius r centred at zero! You may have been thinking that the sum was over the points in the circle? It's just a sum over the individual coordinates to give us the circle formula in the form (coordinate - centre point)^2 < radius Hope that helps, also I can see other questions you have asked in my emails but then I can't find those comments on TH-cam to reply to so you might want to re comment any outstanding questions you have! Many thanks for keep watching my videos and asking questions I really appreciate it!!
@@WHYBmaths " You may have been thinking that the sum was over the points in the circle" Sure. not a big deal i just have two more questions i post it here thank
Gary Tze Hay Lau 3 days ago (edited) just finish this great lectures,i understand all except ONE little question: you said any sets U is in topology should have B(ui) and this B(ui) is a subset of this large U. is it possible to have some S which are closed and still inside this Standard topology ? is that (logically ) possible to happen? If my question is not clear,i might ask in this way: Does every tau i (which is element of topology) should be open set if Topology itself is standard one Because all open set must be included in this standard topology doesnt indicate that this topology includes only open sets... one extra problem: The (discrete)topology = { P(M), phi} seems to be unclear i think The discrete topology itself should be equal to P(M) itself as P(M) itself is a set of set ,and you let topology to become a set of set of set then M is not an element of this topology (M is an element of P(M) where P(M) is an element of this discrete topology) therefore it seems there is logical problem i you write like this? *if M is not is a set (rather than a set of P(M)) in this topology,then this is not called a topology ------------------- -------------------
@@garytzehaylau9432 Yes every set in the topology must be open (in the Hausdorff case), since by definition any closed set cannot contain an open subset around it's boundary points and hence cannot be in the topology. Furthermore since union and intersection of topology elements must also be in the topology we have no way to produce closed sets inside the topology. We can obviously still produce closed sets but they can't be in the topology Regarding P(M) and the discrete topology, it might feel like the discrete topology just has two elements but really P(M) should be understood as shorthand for an entire collection of sets, so the discrete topology has P+1 elements if P is the number of possible subsets of M. The set M is included in this collection and we encounter no difficulties. Just some other minor details to keep in mind, the elements of sets can be treated as sets themselves, so when we write P(M) this can be considered shorthand for a set of all subsets, so the elements of P(M) are themselves sets and these are all in the topology. Essentially this is just notational convenience, rather than writing Tau = {0, {P(M)}} we just write {0, P(M)}. I see what you are saying however about the discrete topology and why we don't just let this be equal to P(M), since this would by necessity contain 0 and M. The actual answer I'm afraid I don't know! I just trust the mathematicians who probably do know and decided it should be this way! I certainly will have a look into this further if I do get the chance. My only guess would be that since the empty set can be considered to be a subset of any set, it is required to be a separate part of the topology for this reason. If anyone has any further insight into this then please do comment!
@@WHYBmathsget it .Very clear except two little one: i can actually draw a closed subset in the region inside U. for the set U inside R^2,i can draw lot of open subset /ball as i like,moreover i could actually draw lot of closed set inside R^2 as well. By defination it said all OPEN SETS MUST BE INCLUDED IN THE TOPOLOGY. (all open sets must be included in the topology doesnt implies all sets in the topology are open) if you have a open set U (which means all element can have a little open ball around and inside U),it is certainly inside this Topology. however we can make the same open set U and includes the boundary (therefore which is a closed set which called S) then we can let this Topology to include this U and S inside at the same time T: { ........,U,S,..............} S and U are actually different elements in this T. since according to the definition of the Topology,we know there is no requirement of the openness{which can seen from the (1)(2)(3) properties of the topology).Now we add one extra property which said (4)"all open sets in R^2 must be included this Topology". "any closed set cannot contain an open subset around it's boundary points" An closed set didn't violate (1)(2)(3) properties of topology(it required sets only) and this is not violated (4) as well since all open sets in this topology doesn't implies the topology doesn't include only open sets you can have a intersection of two closed sets ( S1 and S2)-->you will get a new closed set S3 (S1 intersected with S2) which satisfied (1)(2)(3) and this S3 actually violate "the topology only includes open sets(which is other assumption and we can say it is (5)" and it doesn't violate "all open sets must be included this Topology(4)"" (as no mention of the closed set in this argument) therefore sometimes i get little confused because of the logical structure you presents in lecture #3
I love your videos. One question, for g to be a metric shouldn't it also have a 3rd property, satisfying the triangular inequality? g(x,y) + g(y,z) >= g(x,z)
Yes I forgot to mention this property! I was going to address this when I make a video about the metric tensor however, but thanks for mentioning it here!
Amazing videos. Thanks for sharing.
Good explanation
👏🏻👏🏻👏🏻👏🏻👏🏻
do you live near an airport. nice video!
Great series of videos! A couple of questions:
1.) The metric operates on a pair of vectors, correct? But we haven't endowed our sets here with a vector property as far as I can tell. Or is that implicit in defining a topological space?
2.) Does the metric always have to map to the reals? Or can more exotic metrics be constructed that map to e.g. C or N?
3.) Won't the Hausdorff property always be true unless you somehow include two copies of a point in your set? Struggling to understand the intuition behind this definition.
1) So I may have gotten ahead of myself here and begun referring to the metric tensor by mistake, which does act on a pair of vectors to give a notion of `distance', however already relying on additional structure that we don't yet have (vectors). The metric in this case simply refers to a function of pairs of points in the space.
2) No, more exotic metrics could easily be considered but are not particularly useful for our case working over R, indeed, the case of a discrete topology would have to have a metric in N!
3)We usually would like Hausdorff to be true, at least for everything that would be considered `standard' and for everything else following (manifolds). The intuition is pretty obscured unless you really dig deep into discussing limits and continuity, so for our purposes we can just regard it as a `separation axiom' that effectively lets us talk about the points of a topological space as being `unique', in the sense that they are all surrounded by their own open balls that do not intersect.
Sorry I've taken so long to reply, hope that is somewhat helpful feel free to ask more!
Can the metric also map to non real spaces like say the complex space?
In principle I don't see why not, although I'm not familiar with any cases where this is used, since we need to have the positive definite condition, allowing for complex values can complicate this considerably.
i finish this lecture. i get confused (but not very ) when you represents the open ball with x i and y i (in lecture 3 )
since there is a open ball for only one centre
if you represents your xi in R^2 (lecture #3) or R ,why do you write something like :
B(xi) = { yi : sqrt ( sum i ( xi -yi )^2) < r }
i think xi should be only c or xj becuase if you sum over the yi- xi
then you are not summing only one xi only ? ( not a lot of yi around one xi at the centre but rather a pair of yi and xi )
i think the "sum" terms can be omitted since there is no point to sum up all yi in the argument
for all yi such that : |yi(an element in R^2) - x (an centre element in R^2)|
Yes you are right it is alot easier to just use the modulus definition to define the ball. Maybe I was unclear in how I used the sum symbol however and should have used separate indices, the indices here are only referring to the particular coordinate in R^n. It might have been better to just call the centre of the circle c rather than x_i, but I gave it an index so that it's clear each component of the centre is subtracted from the correct component of the y_i coordinate
If the sum formula is unclear just consider the case for R^2 and let x_i={0,0}. If our coordinates for the points in the ball are y_i=(y1,y2) then B(0,0) = sqrt[sum(y_i -x_i)^2] = sqrt[(y1 - 0)^2+(y2 - 0)^2] < r
Which is the circle of radius r centred at zero! You may have been thinking that the sum was over the points in the circle? It's just a sum over the individual coordinates to give us the circle formula in the form (coordinate - centre point)^2 < radius
Hope that helps, also I can see other questions you have asked in my emails but then I can't find those comments on TH-cam to reply to so you might want to re comment any outstanding questions you have! Many thanks for keep watching my videos and asking questions I really appreciate it!!
@@WHYBmaths " You may have been thinking that the sum was over the points in the circle"
Sure. not a big deal
i just have two more questions
i post it here thank
Gary Tze Hay Lau
3 days ago (edited)
just finish this great lectures,i understand all except ONE little question:
you said any sets U is in topology should have B(ui) and this B(ui) is a subset of this large U.
is it possible to have some S which are closed
and still inside this Standard topology ?
is that (logically ) possible to happen?
If my question is not clear,i might ask in this way:
Does every tau i (which is element of topology) should be open set if Topology itself is standard one
Because all open set must be included in this standard topology doesnt indicate that this topology includes only open sets...
one extra problem:
The (discrete)topology = { P(M), phi}
seems to be unclear
i think The discrete topology itself should be equal to P(M) itself
as P(M) itself is a set of set ,and you let topology to become a set of set of set
then M is not an element of this topology (M is an element of P(M) where P(M) is an element of this discrete topology) therefore it seems there is logical problem i you write like this?
*if M is not is a set (rather than a set of P(M)) in this topology,then this is not called a topology
-------------------
-------------------
@@garytzehaylau9432 Yes every set in the topology must be open (in the Hausdorff case), since by definition any closed set cannot contain an open subset around it's boundary points and hence cannot be in the topology. Furthermore since union and intersection of topology elements must also be in the topology we have no way to produce closed sets inside the topology. We can obviously still produce closed sets but they can't be in the topology
Regarding P(M) and the discrete topology, it might feel like the discrete topology just has two elements but really P(M) should be understood as shorthand for an entire collection of sets, so the discrete topology has P+1 elements if P is the number of possible subsets of M. The set M is included in this collection and we encounter no difficulties. Just some other minor details to keep in mind, the elements of sets can be treated as sets themselves, so when we write P(M) this can be considered shorthand for a set of all subsets, so the elements of P(M) are themselves sets and these are all in the topology. Essentially this is just notational convenience, rather than writing Tau = {0, {P(M)}} we just write {0, P(M)}.
I see what you are saying however about the discrete topology and why we don't just let this be equal to P(M), since this would by necessity contain 0 and M. The actual answer I'm afraid I don't know! I just trust the mathematicians who probably do know and decided it should be this way! I certainly will have a look into this further if I do get the chance. My only guess would be that since the empty set can be considered to be a subset of any set, it is required to be a separate part of the topology for this reason. If anyone has any further insight into this then please do comment!
@@WHYBmathsget it .Very clear except two little one:
i can actually draw a closed subset in the region inside U.
for the set U inside R^2,i can draw lot of open subset /ball as i like,moreover i could actually draw lot of closed set inside R^2 as well.
By defination it said all OPEN SETS MUST BE INCLUDED IN THE TOPOLOGY. (all open sets must be included in the topology doesnt implies all sets in the topology are open)
if you have a open set U (which means all element can have a little open ball around and inside U),it is certainly inside this Topology.
however we can make the same open set U and includes the boundary (therefore which is a closed set which called S)
then we can let this Topology to include this U and S inside at the same time
T: { ........,U,S,..............}
S and U are actually different elements in this T.
since according to the definition of the Topology,we know there is no requirement of the openness{which can seen from the (1)(2)(3) properties of the topology).Now we add one extra property which said (4)"all open sets in R^2 must be included this Topology".
"any closed set cannot contain an open subset around it's boundary points"
An closed set didn't violate (1)(2)(3) properties of topology(it required sets only)
and this is not violated (4) as well since all open sets in this topology doesn't implies the topology doesn't include only open sets
you can have a intersection of two closed sets ( S1 and S2)-->you will get a new closed set S3 (S1 intersected with S2) which satisfied (1)(2)(3)
and this S3 actually violate "the topology only includes open sets(which is other assumption and we can say it is (5)" and it doesn't violate "all open sets must be included this Topology(4)"" (as no mention of the closed set in this argument)
therefore sometimes i get little confused because of the logical structure you presents in lecture #3
I missed your dog