Purchase Solution

Disjoint Closed Point Sets

Not what you're looking for?

Ask Custom Question

Prove:

If H and K are disjoint closed point sets, then there exist open point sets U and V containing H and K respectively such that cl(U) and cl(V) are disjoint.

Purchase this Solution
Solution Summary

This solutions contains a proof regarding disjoint closed point sets. It is solved in an attached Word document.

Purchase this Solution
Free BrainMass Quizzes
Probability Quiz

Some questions on probability

Exponential Expressions

In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.

Solving quadratic inequalities

This quiz test you on how well you are familiar with solving quadratic inequalities.

Multiplying Complex Numbers

This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.

Geometry - Real Life Application Problems

Understanding of how geometry applies to in real-world contexts

View More Free Quizzes
Related BrainMass Solutions

  • Covering Spaces : Compact Hausdorff Spaces and Homomorphisms

    If s is compact subspace in a hausdorff space X , it is closed. Proof:1 Consider any point x not in s. For each y in s, embed y in an open set disjoint from an open set containing x. Do this for each y in s, and we have an open cover.

  • Real Analysis : Connected and Disconnected Sets

    Since B is the closure of A, then B is a closed set and any limit point of B belongs to B. This means B' is contained in B. Conversely, if x is a point in B, I claim x is a limit point of B. Now we have two cases.

  • Closed set proof

    Now I use the other two equivalent definition of closed set to prove this problem. 1. The complement of a closed set is an open set. 2. Each point outside of a closed set has a disjoint neighborhood from this closed set.

  • Real Analysis : Baire Category Theorem

    Also x1 is compact, and a compact set in a hausdorff space is closed, so x1 is closed. These are two distinct closed sets in c1. since c1 is both compact and hausdorff, it is normal. This means x1 and b1 can be separated in disjoint open sets.

  • Co-Finite Topology

    If X is not connected, there are two disjoint nomempty open sets A, B such that X = A union B. Then XA and XB are countable sets and X = (XA) union (XB), therefore X is countable.

  • Topology of Surfaces: Point-Set Topology in R^n.

    Prove that X is connected if and only if X cannot be written as a union of two on-empty disjoint sets which are open relative to X. PROBLEM (Exercise 2.32). Let f : R^n → R^n be a continuous function.

  • Question about Connected point sets

    This is a proof regarding connected point sets.

  • Topology : Connected Spaces and Explosion Point

    Proof: Since , and are two nonempty disjoint open sets. Then and are disjoint. So and are disjoint. implies that . Since is disconnected, is open, then is not open.

  • discrete topology

    363362 discrete topology covering map A covering map is a map p: E -> B with the property that each point b, an element of B, has a neighborhood U such that p^-1(U) is a disjoint union of open sets V_alpha such that, for each alpha, the restriction of

View More Related Solutions