Chapter 3: Problem 9
If \(A=\\{x:\|x\|<1\\}\) and \(B=\\{x:\|x\| \leq 1\\}\) show that \(\bar{A}=B\)
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Chapter 3: Problem 9
If \(A=\\{x:\|x\|<1\\}\) and \(B=\\{x:\|x\| \leq 1\\}\) show that \(\bar{A}=B\)
These are the key concepts you need to understand to accurately answer the question.
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Prove that \(\mathbb{Z}\) (the integers) is a closed subset of \(\mathbb{R}\).
Let \(f\) be a continuous mapping from a normed linear space \(X\) into
\(\mathbb{R}\). Then the following sets are open:
$$
\\{x: f(x)<\alpha\\}, \quad\\{x: f(x)>\beta\\}, \quad\\{x:
\alpha
Let \(A\) be a subset of the normed linear space \(X\) satisfying \(\lambda a \in A\) whenever \(a \in A\) and \(\lambda \geq 0\). Show that \(A\) is closed if and only if \(A \cap\\{x:\|x\| \leq 1\\}\) is closed.
Show that if \(G\) is open in the normed linear space \(X\) and \(A\) is a set with \(A \cap G\) empty, then \(\bar{A} \cap G\) is also empty.
Show that $$ \operatorname{int}(\operatorname{int} A)=\operatorname{int} A \quad \text { and } \quad \operatorname{int}(A \cap B)=\operatorname{int} A \cap \operatorname{int} B $$ Give an example to show that \(\operatorname{int}(A \cup B)=\operatorname{int} A \cup \operatorname{int} B\) is not necessarily true.
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