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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 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.

Show that the set \(V^{\prime}\) consisting of bounded linear functionals on a Banach space \(V\) is a normed vector space with respect to the norm $$ \|\varphi\|=\sup [M|| \varphi(x) \mid \leq M\|x\| \text { for all } x \in V \mid $$ Show that this norm is complete on \(V^{\prime}\).

Let \(X\) and \(Y\) be topological spaccs and \(f . X \times Y \rightarrow X\) a continuous map. For each fixed \(a \in X\) show that the map \(f_{n}: Y \rightarrow X\) defined by \(f_{\theta}(v)=f(a, v)\) is contunous.

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 \(0 \in V\).

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 \(\|\cdot\|_{1}\).
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