I have failed the DIE exam, and instead of prepairing for reexamination, i watch your vids every night because they are much more interesting. Thanks a lot!
Thanks, I'm happy you are enjoying them! I have covered about a 10-20% of the material I intend to present in the 5 videos so far. So I will probably have to make at least 25 more. I don't have it planned out in advance, so it is hard to say.
Just a technical point about your proof at 38:15: what you have shown is that preimg(U)nV is open in V as a subspace of X. The reason it is also open in X is that V is open in X, and in general if A is open in (subspace) B, and B is open in C, then A is open in C (I.e. openness is subspace transitive). But this lemma also deserves proof. You would have to define the subspace topology.
Yes, you are right that if we interpret the continuity of the restricted maps with respect to the subspace topology (as is natural), we would need to argue that open subsets of open subspaces are again open. The argument as it stands could be saved by interpreting the continuity of the restricted map as saying that preimages of open sets under the restricted map are open in X. In our case, the two definitions are equivalent because of the openness of the subspaces considered. But the second definition would be wrong for restricting to more general subspaces. I don't say exactly what I mean by the continuity of the restriction, probably because I did not want to talk about subspaces at that time.
Thanks for your lecture. I love your video. But maybe you can use more steps to help our understanding: f(V) is open in Y ( V is open in X and X->Y is continuous in V ), and U is also open in Y, then f(V) ∩ U is open in Y, then V ∩ f-1(U) ( the preimage of f(V) ∩ U ) is open in X ( the continuity again ).
Great lecture! I've been following it very closely and had one doubt. Would be grateful if someone could help. At 37:41, you prove that f(|V) ^(-1) (U) i.e. the purple set is open because it is the pre-image of U (an open set) under a restrictive function f which you assumed to be continuous. I'm assuming you draw this conclusion from the proposition (c) proved at 24:32 If so, doesn't this require f^(-1) (U) to be an open set first (as mentioned in the proposition) ? Since it is not an iff statement, we can't draw any reverse conclusions. Thanks
Thanks for all the positive feedback! I'm very happy you are enjoying the videos. There are many other topics I would love to cover. High on the list are Algebra and then Algebraic Topology. However it takes me around 6 hours to produce a video of this quality, so I'm afraid I will always be behind what I would like to cover.
Can you show how continuity would breakdown if one open set had a closed pre-image or here if one closed set had an open preimage? There must be a problem of convergence on the limit of the function. These lectures are awesome. Thank you.
Remember that "closed" is does not mean not open. Rather it means that the complement of the set in question is open. Hence it is possible for the preimage of an open set to be closed, even if the function is continuous. In this case the preimage would be both open and closed. As an example of such a continuous map, consider the constant map R -> R which maps each x in R to 0. Then the preimage of each open set is either empty or all of R. Both of these sets are both open and closed in R. Regarding the preimages of closed sets the same thing holds. If a function is continuous, then the preimages of all closed sets are closed, but they might also be open at the same time. In the specific case of R, the only sets that are both open and closed are R and the empty set. This follows from the fact that R is connected. Hence in R (or any connected space) you can indeed infer that a set cannot be open and closed at the same time if it is nonempty and not all of R.
You talk about balls in a general topological space (X, T), could you define a ball supposing the topology is not induced by a metric? I head your answer at 10:16, now it's clear.
Didn't follow the argument about continuity that well. I was kind of expecting a measure to be introduced but I may be too stuck in my metric space ways.
What does so called restricted map f : U -> Y (where U is a subset of set X) mean? Why isn't it a map f: U -> V where U is a subset of X and U is a subset of Y? how could f map points of U to the whole set Y?
The notation f: X -> Y means we assign every x in X a point f(x) in Y. Hence the image of f does not need to cover the entire set Y (i.e be surjective). The restricted function f: U -> Y is defined by sending every u in U to f(u) in Y. Again, it does not cover all of Y. As an example, think of the squaring function (-)^2 which sends x to x^2. We can think of (-)^2 : R -> R as a function from the reals R to R even though not every real number is the square of some number. We can also restrict the domain of this function to for instance view it as a function [0,1] -> R. Restricting the codomain of a function is also possible, but one needs to be careful that one doesn't remove points that the function assumes. In fact, by restricting the codomain of any function, you can make it surjective.
The way we have defined continuity (preimages of open sets are open) makes it a global property that a function either has or does not have. In analysis one usually defines what it means for a function to be continuous at a point and then says a function is continuous if it is continuous at every point. In that setting it is thus possible to be continuous at some points but no at others. According to our definition, these functions would not be continuous. One way to easily get examples of discontinuous functions is to define them by cases. For instance the function f: R -> R that sends x to -1 if x = 0 is not continuous. The reason is that the preimage of a small neighborhood containing +1, but not -1 is the interval [0,\infinity) which is not open in R.
This deserves more views.
Gracias men, me hiciste entender mucho sobre este tema, que lo tenía como mi nemesis
I have failed the DIE exam, and instead of prepairing for reexamination, i watch your vids every night because they are much more interesting. Thanks a lot!
You are welcome! I'm happy you are enjoying the videos.
Great topology lessons, thank you very much! How many videos do you plan to make?
Thanks, I'm happy you are enjoying them! I have covered about a 10-20% of the material I intend to present in the 5 videos so far. So I will probably have to make at least 25 more. I don't have it planned out in advance, so it is hard to say.
Very informative! It helped me review many relevant important concepts. Thank you very much!
You are very welcome!
Just a technical point about your proof at 38:15: what you have shown is that preimg(U)nV is open in V as a subspace of X. The reason it is also open in X is that V is open in X, and in general if A is open in (subspace) B, and B is open in C, then A is open in C (I.e. openness is subspace transitive). But this lemma also deserves proof. You would have to define the subspace topology.
Yes, you are right that if we interpret the continuity of the restricted maps with respect to the subspace topology (as is natural), we would need to argue that open subsets of open subspaces are again open.
The argument as it stands could be saved by interpreting the continuity of the restricted map as saying that preimages of open sets under the restricted map are open in X. In our case, the two definitions are equivalent because of the openness of the subspaces considered. But the second definition would be wrong for restricting to more general subspaces. I don't say exactly what I mean by the continuity of the restriction, probably because I did not want to talk about subspaces at that time.
Thanks for your lecture. I love your video. But maybe you can use more steps to help our understanding: f(V) is open in Y ( V is open in X and X->Y is continuous in V ), and U is also open in Y, then f(V) ∩ U is open in Y, then V ∩ f-1(U) ( the preimage of f(V) ∩ U ) is open in X ( the continuity again ).
Great lecture! I've been following it very closely and had one doubt. Would be grateful if someone could help.
At 37:41, you prove that f(|V) ^(-1) (U) i.e. the purple set is open because it is the pre-image of U (an open set) under a restrictive function f which you assumed to be continuous.
I'm assuming you draw this conclusion from the proposition (c) proved at 24:32
If so, doesn't this require f^(-1) (U) to be an open set first (as mentioned in the proposition) ? Since it is not an iff statement, we can't draw any reverse conclusions.
Thanks
Beautiful lecture. The proofs are wonderful. Would you consider making a more advanced playlist? Thank you!!!
Thanks for all the positive feedback! I'm very happy you are enjoying the videos. There are many other topics I would love to cover. High on the list are Algebra and then Algebraic Topology. However it takes me around 6 hours to produce a video of this quality, so I'm afraid I will always be behind what I would like to cover.
@@mariusfurter Algebraic topology is really badly presented please consider offering a playlist. Thank you again!!!
Can you show how continuity would breakdown if one open set had a closed pre-image or here if one closed set had an open preimage? There must be a problem of convergence on the limit of the function. These lectures are awesome. Thank you.
Remember that "closed" is does not mean not open. Rather it means that the complement of the set in question is open. Hence it is possible for the preimage of an open set to be closed, even if the function is continuous. In this case the preimage would be both open and closed. As an example of such a continuous map, consider the constant map R -> R which maps each x in R to 0. Then the preimage of each open set is either empty or all of R. Both of these sets are both open and closed in R.
Regarding the preimages of closed sets the same thing holds. If a function is continuous, then the preimages of all closed sets are closed, but they might also be open at the same time.
In the specific case of R, the only sets that are both open and closed are R and the empty set. This follows from the fact that R is connected. Hence in R (or any connected space) you can indeed infer that a set cannot be open and closed at the same time if it is nonempty and not all of R.
You talk about balls in a general topological space (X, T), could you define a ball supposing the topology is not induced by a metric? I head your answer at 10:16, now it's clear.
Didn't follow the argument about continuity that well. I was kind of expecting a measure to be introduced but I may be too stuck in my metric space ways.
What does so called restricted map f : U -> Y (where U is a subset of set X) mean? Why isn't it a map f: U -> V where U is a subset of X and U is a subset of Y? how could f map points of U to the whole set Y?
The notation f: X -> Y means we assign every x in X a point f(x) in Y. Hence the image of f does not need to cover the entire set Y (i.e be surjective). The restricted function f: U -> Y is defined by sending every u in U to f(u) in Y. Again, it does not cover all of Y. As an example, think of the squaring function (-)^2 which sends x to x^2. We can think of (-)^2 : R -> R as a function from the reals R to R even though not every real number is the square of some number. We can also restrict the domain of this function to for instance view it as a function [0,1] -> R.
Restricting the codomain of a function is also possible, but one needs to be careful that one doesn't remove points that the function assumes. In fact, by restricting the codomain of any function, you can make it surjective.
Thank you very much! @@mariusfurter
How could I write a function that is not continuous?
Can a function be continuous and not continuous at the same time?
The way we have defined continuity (preimages of open sets are open) makes it a global property that a function either has or does not have. In analysis one usually defines what it means for a function to be continuous at a point and then says a function is continuous if it is continuous at every point. In that setting it is thus possible to be continuous at some points but no at others. According to our definition, these functions would not be continuous.
One way to easily get examples of discontinuous functions is to define them by cases. For instance the function f: R -> R that sends x to -1 if x = 0 is not continuous. The reason is that the preimage of a small neighborhood containing +1, but not -1 is the interval [0,\infinity) which is not open in R.