Thanks, I'm happy to hear you are enjoying the videos. Yes, I would like to cover those topics eventually, but I will probably do so in a separate series on algebraic topology.
Nice video! A small note, in the proof that if X is locally path-connected, then components are identical to path components (around 57:40), I don't think it is necessary to show that the component C itself is open (even though that is true). We just need that P is contained in C and that P is both open and closed in X. These are sufficient to show that P is both open and closed in C (under the subspace topology that it inherits from X).
Also, it's not true in general that path components are closed (55:15). A good counterexample is TSC where there are two path components. The sine part of the curve is not closed, since its complement (the vertical segment) is not open.
Amazing lecture. Thanks. I was watching this wondering how to show that a nonempty closed open and connected subset must be the component of its elements and realized that it would cause a contradiction to assume otherwise.
Hey, Really love your content. Kindly extend this lecture series to Fundamental Groups and homotopy.
Thanks, I'm happy to hear you are enjoying the videos. Yes, I would like to cover those topics eventually, but I will probably do so in a separate series on algebraic topology.
Nice video! A small note, in the proof that if X is locally path-connected, then components are identical to path components (around 57:40), I don't think it is necessary to show that the component C itself is open (even though that is true). We just need that P is contained in C and that P is both open and closed in X. These are sufficient to show that P is both open and closed in C (under the subspace topology that it inherits from X).
Also, it's not true in general that path components are closed (55:15). A good counterexample is TSC where there are two path components. The sine part of the curve is not closed, since its complement (the vertical segment) is not open.
Amazing lecture. Thanks. I was watching this wondering how to show that a nonempty closed open and connected subset must be the component of its elements and realized that it would cause a contradiction to assume otherwise.
55:15 one should show that P is closed.
Each connected component is only closed need not be open??