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Chapter 10: Problem 11

A topological space \(X\) is called normal if for evcry pair of disjo?nt closcd suhsets \(A\) and \(B\) there exist disjoint open sets \(U\) and \(V\) sich that \(A \subset U\) and \(B \subset V\) Show that every metric space is nomal

Short Answer

Expert verified
Every metric space is a normal space as for any pair of disjoint closed sets in the metric space, it is possible to find disjoint open sets containing them.

Step by step solution

01

Clarify Definitions

A metric space is a set for which distances between all its pairs of elements are specified. The distance between two elements is a non-negative real number that measures how far apart those elements are. A topological space is normal if, given any two disjoint closed sets, it is always possible to find two disjoint open sets containing them.
02

Consider Disjoint Closed Sets in the Metric Space

Let \(A\) and \(B\) be two disjoint closed sets in the metric space \(X\). We need to show that there exist disjoint open sets \(U\) and \(V\) such that \(A \subset U\) and \(B \subset V\).
03

Construct the Open Sets

For each point \(a \in A\) and \(b \in B\), the distance between \(a\) and \(b\), which we can denote \(d(a, b)\), is positive because \(A\) and \(B\) are disjoint. Let \(r = \frac{1}{2}d(a, b)\). Now, consider the open balls \(B(a, r)\) and \(B(b, r)\) centered at \(a\) and \(b\) respectively with radius \(r\). These open balls are disjoint, and the union of all such balls \(B(a, r)\) for \(a \in A\) is an open set containing \(A\), and similarly, the union of all balls \(B(b, r)\) for \(b \in B\) is an open set containing \(B\), and these two open sets are disjoint.
04

Conclusion

Therefore, for every pair of disjoint closed sets in the metric space, we can find disjoint open sets containing them, hence proving that every metric space is a normal space.

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Most popular questions from this chapter

Show that a linear map \(T: V \rightarrow W\) between topological vector spaces is continuous everywhere on \(V\) if and only if it is continuous at the origin \(0 \in V\).

Show that the following are all norms in the vector space \(\mathbb{R}^{2}\) : $$ \begin{aligned} &\|\mathbf{u}\|_{1}=\sqrt{\left(u_{1}\right)^{2}+\left(u_{2}\right)^{2}} \\ &\|\mathbf{u}\|_{2}=\max \left[\left|u_{1}\right|,\left|u_{2}\right|\right\\} \\\ &\|\mathbf{u}\|_{3}=\left|u_{1}\right|+\left|u_{2}\right| \end{aligned} $$ What are the shapes of the open balls \(B_{a}(\mathrm{u})\) ? Show that the topologes generated by these norms are the same.

We say two norms \(\|u\|_{1}\) and \(\|u\|_{2}\) on a vector space \(V\) are equivalent if there exist constants \(A\) and \(B\) such that $$ \|u\|_{1} \leq A\|u\|_{2} \quad \text { and }\|u\|_{2} \leq B\|u\|_{1} $$ for all \(u \in V\). If two norms are equivalent then show the following: (a) If \(u_{n} \rightarrow u\) with respect to one norm then this is also true for the other norm. (b) Every linear functional that is continuous with respect to one norm is continuous with respect to the other norm. (c) Let \(V=C[0,1]\) be the vector space of continuous complex functions on the interval \([0,1]\). By considering the sequence of functions $$ f_{n}(x)=\frac{n}{\sqrt{\pi}} \mathrm{e}^{-m^{2} x^{2}} $$ show that the norms $$ \|f\|_{1}=\sqrt{\int_{0}^{1}|f|^{2} \mathrm{~d} x} \text { and }\|f\|_{2}=\max \\{f(x)|| 0 \leq x \leq 1\\} $$ are not equivalent. (d) Show that the linear functional defined by \(F(f)=f(1)\) is contunuous wath respect to \(\|\cdot\|_{2}\) but not with respect to \(\|\cdot\|_{1}\).

If \(G_{0}\) is the component of the identity of a locally connected topological group \(G\), the factor group \(G / G_{0}\) is called the group of components of \(G .\) Show that the group of components is a discrete topological group with respect to the topology induced by the natural projection map \(\pi: g \mapsto g G_{0}\)

Let \(V\) be a Banach space and \(W\) te a vector subspace of \(V .\) Define its closture \(\bar{W}\) to be the union of \(W\) and all hmuts of Cauchy sequences of elements of \(W\). Show that \(\bar{W}\) is a closed vector subspace of \(V\) in the sense that the limit points of all Cauchy scquences in \(\bar{W}\) lie in \(\bar{W}\) (note that the Cauchy sequences may include the newly added limit points of \(W\) ).
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