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Chapter 10: Problem 5

If \(A\) is a dense set in a topological space \(X\) and \(U \subseteq X\) is open, show that \(U \subseteq \overline{A \cap U}\).

Short Answer

Expert verified
The given statement \(U \subseteq \overline{A \cap U}\) is proved using the definitions of dense set, open set, and closure. Our proof concludes that for every \(x \in U\), we have \(x \in \overline{A \cap U}\), which shows \(U \subseteq \overline{A \cap U}\).

Step by step solution

01

Analyzing \(U \subseteq \overline{A \cap U}\)

We aim to show that every element x of U is also an element of \(\overline{A \cap U}\), where \(\overline{A \cap U}\) denotes the closure of the set \(A \cap U\). The expression \(A \cap U\) represents intersection of set \(A\) and \(U\), so the points of \(A \cap U\) are the points which are common to both \(A\) and \(U\).
02

Utilizing the definition of a dense set

Since \(A\) is given as a dense set in \(X\), by its definition, the smallest closed set containing \(A\) is the entire topological space \(X\). Therefore, this includes the open set \(U\). Hence, the closure of set \(A\), \(\overline{A}\), contains \(U\), that is \(U \subseteq \overline{A}\).
03

Utilizing the definition of closure

The definition of the closure of a set includes all limit points of the set, i.e., for a point to be in the closure of a set, every open set containing the point must intersect the set. So the points in \(\overline{A \cap U}\) intersect \(A\) and \(U\), meaning that if \(x \in U\), then \(x \in \overline{A \cap U}\).
04

Concluding the proof

Bringing these steps together, we have shown that for every \(x \in U\), \(x \in \overline{A \cap U}\). Hence, \(U \subseteq \overline{A \cap U}\), which completes the proof as required.

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Most popular questions from this chapter

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

Show that cvery space \(F(S)\) is complete with respect to the supremum norm of Example 10.26. Hence show that the vector space \(\ell_{\infty}\) of bounded infinite complex sequences is a Banach space with respect to the norm \(\|\mathrm{x}\|=\sup \left(x_{t}\right)\).

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 \(0 \in V\).

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 \(\pi: g \mapsto g G_{0}\)

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.
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