The topological definition of continuity is deceptively simple and it takes quite a while to wrap one's head around why it corresponds to (or in fact generalizes) the analytical definition. My favorite way to understand it is to consider the topological definition of a not-continuous function: there exists an open set U such that f^(-1)(U) is not open. For a set to be not open, it has to contain a point which is not an interior point. This not-interior point can be thought of as a point of discontinuity, and lends itself to drawing a picture. We then think of continuity as lacking any such points of discontinuity.
By regarding the definition you gave for the limit and for the continuity then the existence of a limit of a function at a point in which it's defined and the continuity of the same fuction at this point are equivalent.
I am confused at 4:28 We should have taken the intersection of delta neighborhood and U there, but if we do that then proving delta neighborhood inside 'U' is not that way😢.
Good question. This is not a popular opinion, but I think that lower level math classes (calculus/linear algebra) are sometimes too focused on the needs of non-math majors. Perhaps it has something to do with this. Other departments are not continuously expected to make their subject applicable to every other field of study...
Can an open set be a finite segment of the real line? It seems like you could take a point in the segment then find a point smaller in the neighborhood of that point then find a point smaller than that point in its neighborhood and repeat off to infinity... Or does not including the exact endpoints allow that and still limit the length?
I’m not sure if I understood your question correctly; however, if that is your question, an open subset of the real numbers can indeed be finite - just take the empty set, which is both finite and open. However, the empty set is the only finite open set under the usual Euclidean topology of the reals. This is because finite seats are compact; recall that topologists say a set is compact if every open cover of that set has a finite subcover. Of course, any finite set trivially has this property, and so it is compact. However, in the real numbers (and, more generally, in any Hausdorff space), all compact sets are closed, which means that the set in question must be both open and closed - of course, the empty set and the entire real number line have this property, but there are actually no other subsets of the reals that are both closed and open (this is, I hope, intuitively clear, but it’s not completely trivial to prove - if you get stuck, I can write down a proof of it), and so the empty set is the only finite open subset of the reals.
Beatorīche Ah, thanks a lot. So it is different than the notion of open that’s indicated by writing, say, (3..5) rather than [3..5], that’s what I was confused about.
@@Reliquancy I just want to comment that you're right to think that the open interval (3,5) is an open subset of the real line and [3,5] is a closed subset. If I've understood what you were saying correctly, you're also right that for example 4 is in (3,5), 3.5 is in (3,5), 3.25 as well... however the limit of this sequence is 3 which is not in (3,5). When this kind of thing happens people say 3 is a limit point of the set. In general, the smallest closed set containing an open set U is U together with its limit points, which maybe makes some intuitive sense
For reference, all open sets in the real numbers can be written by taking arbitrary unions and/or finite intersections of intervals of the form (a, b) where a and b are rationals. An open interval in the reals thus satisfies the above.
The topological definition of continuity is deceptively simple and it takes quite a while to wrap one's head around why it corresponds to (or in fact generalizes) the analytical definition. My favorite way to understand it is to consider the topological definition of a not-continuous function: there exists an open set U such that f^(-1)(U) is not open. For a set to be not open, it has to contain a point which is not an interior point. This not-interior point can be thought of as a point of discontinuity, and lends itself to drawing a picture.
We then think of continuity as lacking any such points of discontinuity.
9:50
Anyway, have a good rest of the morning, afternoon, evening, wherever you are.
Notice that the preimage is not the same as the inverse.
The inverse of a function only exists if it's a bijection, while the preimage always exists.
You are very good teacher! You explain so good! Thank you!
It’s actually sometimes hard to find the intuition to this theorem but I finally understand it intuitively.
By regarding the definition you gave for the limit and for the continuity then the existence of a limit of a function at a point in which it's defined and the continuity of the same fuction at this point are equivalent.
I am confused at 4:28
We should have taken the intersection of delta neighborhood and U there, but if we do that then proving delta neighborhood inside 'U' is not that way😢.
I wonder why, apart from possible historical reasons, Topology & Calculus are not taught in conjunction.
Good question. This is not a popular opinion, but I think that lower level math classes (calculus/linear algebra) are sometimes too focused on the needs of non-math majors. Perhaps it has something to do with this.
Other departments are not continuously expected to make their subject applicable to every other field of study...
Thanks! Does the definition generalize to any general Topology?
Yes. A function between two topological spaces is defined to be continuous if for all open subsets of Y the preimage is open in X.
Nice teaching amazing sir
Nice exposition: close points go to close points.
Can an open set be a finite segment of the real line? It seems like you could take a point in the segment then find a point smaller in the neighborhood of that point then find a point smaller than that point in its neighborhood and repeat off to infinity... Or does not including the exact endpoints allow that and still limit the length?
I’m not sure if I understood your question correctly; however, if that is your question, an open subset of the real numbers can indeed be finite - just take the empty set, which is both finite and open. However, the empty set is the only finite open set under the usual Euclidean topology of the reals. This is because finite seats are compact; recall that topologists say a set is compact if every open cover of that set has a finite subcover. Of course, any finite set trivially has this property, and so it is compact. However, in the real numbers (and, more generally, in any Hausdorff space), all compact sets are closed, which means that the set in question must be both open and closed - of course, the empty set and the entire real number line have this property, but there are actually no other subsets of the reals that are both closed and open (this is, I hope, intuitively clear, but it’s not completely trivial to prove - if you get stuck, I can write down a proof of it), and so the empty set is the only finite open subset of the reals.
Beatorīche Ah, thanks a lot. So it is different than the notion of open that’s indicated by writing, say, (3..5) rather than [3..5], that’s what I was confused about.
@@Reliquancy I just want to comment that you're right to think that the open interval (3,5) is an open subset of the real line and [3,5] is a closed subset. If I've understood what you were saying correctly, you're also right that for example 4 is in (3,5), 3.5 is in (3,5), 3.25 as well... however the limit of this sequence is 3 which is not in (3,5).
When this kind of thing happens people say 3 is a limit point of the set. In general, the smallest closed set containing an open set U is U together with its limit points, which maybe makes some intuitive sense
Álvaro L. Martínez Thanks I guess @Beatorīche meant a finite segment as in a finite number of elements, I guess I meant a finite length interval.
For reference, all open sets in the real numbers can be written by taking arbitrary unions and/or finite intersections of intervals of the form (a, b) where a and b are rationals. An open interval in the reals thus satisfies the above.
1:07 contained in me?
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