You're being incredibly rude to other community members. Since this is obviously not OK, would you please stop?

The problem with your posts is not that they are hard to understand, it's that they are not nice to read. For one, you come off as arrogant, as if you were the only one who has any idea what they are talking about. You present a solution to help disadvantaged students learn better, and I agree that your solution is fine. But it will only work for students that have the will, time and ability to help themselves, and I think that is only a tiny minority of all students, and even tinier when you only consider economically disadvantaged ones.

That your first post didn't mention any of the factors that could make your solution unworkable probably contributes to the downvotes, because you come across as somewhat naive, and not in a cute way. I don't think it deserved to be flagged, but it wasn't a good post either.

Your second post is much too repetitive. Repetition is very memorable (you got "Get the book. Read the book. Do the exercises." stuck in my head quite well), but if you don't enrich it with convincing arguments, it just becomes tiresome to read.

Your third post is actually quite reasonable (but still really long, not sure if you could have made it shorter). You clarify that your solution is only for talented students and you mention a bunch of things that would be good for such students to know.

Your fourth post is really bad. You directly attack someone who was trying to help you with stylistic advice and tell them that they didn't understand your post, deny that there is any problem and proceed to repeat yourself. That you got called out by a moderator should give you a hint how far off the mark your writing is.

That said, consider me nerd-sniped! I quite enjoyed doing your exercises, although I had to look up some definitions here and there. Unfortunately, they don't fit in a single comment, so I have to see how many self-replies I am allowed.

Example: In a separable metric space, each closed set is the union of a perfect and a set that is at most countable.

Definition of separability [1]: a topological space is called separable if it contains a countable, dense subset; that is, there exists a sequence of elements of the space such that every nonempty open subset of the space contains at least one element of the sequence.

Defintion of perfect set [2]: a subset of a topological space is perfect if it is closed and has no isolated points.

Let C be a closed set in a separable metric space. Choose the perfect P as C minus the set of C's isolated points I. Since they are isolated, we can assign to each point p in I an open neighborhood N_p without any of the neighborhoods overlapping. Because each neighborhood contains at least one distinct element of the separating set S (is that what you call it?), the cardinality of I is at most that of S, i.e. countable. Thus C is the union of a perfect (P) and set that is at most countable (I).

[1] https://en.wikipedia.org/wiki/Separable_space [2] https://en.wikipedia.org/wiki/Perfect_set

Examle: For positive integer n and the set of real numbers R, show that convex f: R^n --> R is continuous.

Choose an arbitrary point x in R^n and consider a hypercube centered in x. Each face of the hypercube together with the centerpoint x determines a hyperpyramid, and each point p within the hypercube is contained in at least one such hyperpyramid.

Since the hyperpyramid is convex, the point p can be expressed as a convex combination of its vertices, which is linear in the coordinates of p. Due to the convexity of f, there is an upper bound of f(p) that is a convex combination of the values of f on the vertices of the hyperpyramid, using the same coefficients. This means that the upper bound is linear in p as well.

Importantly, the bound is identical on the boundary of two hyperpyramids, since the coefficients will only involve those vertices that are shared between them. This means that f is bounded above by a piecewise linear function, which is tight in x (with the trivial convex combination).

There is a second hyperpyramid to consider, which is formed by p together with the opposite face of the hypercube, and which contains x. By the same convexity argument, there is an upper bound on f(x) that is a convex combination with coefficients linear in p.

The bound can be written as sum_v w_v f(v) + w_p f(p) >= f(x) and rewritten as f(p) >= 1/w_p (f(x) - sum_v w_v f(v)). This is possible because w_p = 0 would imply that x is a convex combination of the v alone and therefore in the hyperface. This bound is no longer linear in p, but still continuous.

Like the upper bound, this bound is identical on the boundary of two hyperpyramids containing p, because only the vertices shared by the opposite faces will be relevant. Additionally, the bound is tight in x as well, where the convex combination is w_p = 1, w_v = 0.

In summary, f is bounded above and below by continuous functions in a neighborhood of x, with the bounds coinciding at x itself. This implies that f must be continuous in x as well.

Example: For positive integer n and the set R of real numbers, for any closed set C in R^n, there exists a function f: R^n --> R so that f(x) = 0 for all x in C, f(x) > 0 otherwise, and f is infinitely differentiable.

I have a hunch that this can be solved similarly to the construction of bump function on a given compact set [4], except the support is an open set and I'm not sure what the "appropriate scaling" should look like.

[4] https://en.wikipedia.org/wiki/Bump_function#Existence_of_bum...

Example: Prove that there are no countably infinite sigma algebras.

Definition of sigma algebra [3]: a σ-algebra (also σ-field) on a set X is a collection Σ of subsets of X that includes the empty subset, is closed under complement, and is closed under countable unions and countable intersections.

Assume a countably infinite sigma algebra does exist. Consider the sets in the algebra that do not have non-empty proper subsets in the algebra. In particular, their intersections with other sets are always empty. If there is a countable infinity of them, they can be mapped to the one-element sets of natural numbers, and their closure under the operations of the sigma algebra is isomorphic to its powerset, which is uncountable.

Therefore there can be only finitely many such sets. Now consider the collection S of sets that remains after removing their (finite) closure under the operations of the sigma algebra. Obviously S is still countably infinite. Since each set A in S has a proper non-empty subset B in the sigma algebra, it also has another: the intersection of the complement of B with A. If those subsets are also in S, they in turn can be split into two non-empty proper subsets.

If this process of binary splitting stops at a finite depth, it creates a binary tree whose leaves are not in S (since they can't be split) and whose inner nodes are the unions of their two children. By induction, this represents each node in the tree as the union of finitely many sets not in S, including the root node A. However, this contradicts the choice of A, which means that the process never stops and there is an infinite path of the tree.

Now consider those nodes that are split off this infinite path, i.e. those that are children of a node on the path, but are not on the path themselves. Since each of them is by construction disjoint from their siblings on the path and their descendants, they form a countably infinite collection of disjoint sets. By the same construction as above, they can be mapped to the one-element sets of natural numbers, which means that their closure is uncountably infinite.

Therefore, the assumption that countably infinite sigma algebras exist is false.

[3] https://en.wikipedia.org/wiki/Sigma-algebra

Example: For triangle ABC, by Euclidean construction, find D on AB and E on BC so that the lengths AD = DE = EC.

Running out of time here. I guess you can do it algebraically: two variables for the placement of D and E, two quadratic equations, get the solutions in terms of square roots, multiplication, division etc., then implement the calculation using compass and straightedge.

Example: For positive integer n and the set of real number R , show that, on any finite intersection of closed half spaces C in R^n, any linear function that is bounded above on C achieves its least upper bound on C.

If C is empty (i.e. the closed half spaces do not overlap), the claim is false, since every real number is an upper bound of every function on C, therefore the least upper bound does not exist.

However, for non-empty C and a linear function f bounded above on C, any x in C induces a lower bound l for the upper bounds, which guarantees the existence of a least upper bound u.

The point where u is achieved can be determined by an iterative procedure involving a set of half-spaces H (initially H_0 is empty), a point x in C (initially x_0 is chosen arbitrarily) and a search direction d (initially d_0 is the gradient of f).

The following invariant is maintained: for any boundary of a half-space in H holds that x is on the boundary and d is parallel to it, and for any x' in C, f(x') > f(x) implies that the dot product d (x' - x) is positive.

This holds for the inital conditions, since H_0 is empty and d_0 is the gradient of f, which means that d_0 x' = f(x') > f(x) = d_0 x is equivalent to d_0 (x' - x) > 0.

If d ever becomes zero, the invariant implies that there are no points x' in C with f(x') > f(x), which means that f(x) = u.

Otherwise, the value of x_{n+1} is chosen from the linear subspace x_n + t d_n, where t in [0, infinity). As this is a closed, convex set, its intersection with C (which is an intersection of closed, convex sets) is itself closed and convex. Since f keeps increasing linearly with t, the upper bound u on f immediately induces an upper bound on t. Summarily, this constrains t to an interval [0, s]. Then x_{n+1} = x_n + s d_n.

This step maintains the invariants, since x moves parallel to all boundaries of sub-spaces in H_n and x_n was contained therein, thus x_{n+1} is also contained in all these boundaries. Because the search direction doesn't change, its invariant is maintained as well.

At this point, there must be a half-space h that "cut off" the line at x_{n+1}, which means that x was not moving in parallel to it. Setting adding h to H_n keeps x_{n+1} contained within all the boundaries, but makes d_n no longer parallel to them. By also setting d_{n+1} to the component of d_n that is parallel to the boundary of h, this can be reinstated.

However, now it must be shown that f(x') > f(x) implies d_{n+1} (x' - x) > 0. From the previous step, it holds that f(x') > f(x) implies 0 < d_n (x' - x) = (d_n - d_{n+1} + d_{n+1}) (x' - x) = (d_n - d_{n+1}) (x' - x) + d_{n+1} (x' - x) where d_n - d_{n+1} is perpendicular to the boundary of h and away from it. For d_{n+1} (x' - x) to be non-positive with the overall result staying positive, (d_n - d_{n+1}) (x' - x) would have to be positive. But this means that x' must be a non-zero distance outside h, and is therefore not in C. The only remaining possibility is that the invariant is maintained.

The above algorithm must terminate in a finite number of steps, since there is only a finite number of half-spaces that can be added to H. Because the algorithm can only terminate by finding an x in C where f(x) = u, this proves that the least upper bound is achieved.