Open, Closed, Compact Sets
Important notions for subsets of real numbers.
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Study Open, Closed, Compact Sets
Next, we define some particular sets and special properties of sets.
Definition 1 (\(\varepsilon\)-neighbourhood). Let \(x\in \mathbb{K}\). Then for \(\varepsilon>0\), the \(\varepsilon\)-neighbourhood of \(x\) is defined by the set \[B_{\varepsilon}(x)=\{y\in \mathbb{K}\,:\,|x-y|<\varepsilon\}.\] A set \(M\subset \mathbb{K}\) is called neighbourhood of \(x\), if there exists some \(\varepsilon>0\) such that \[B_{\varepsilon}(x)\subset M.\]
Example 2.
If \(\mathbb{K}=\mathbb{R}\), then the \(\varepsilon\)-neighbourhood of \(x\in\mathbb{R}\) is given by the interval \[B_{\varepsilon}(x)=(x-\varepsilon,x+\varepsilon).\]
If \(\mathbb{K}=\mathbb{C}\), \(\varepsilon>0\), then the \(\varepsilon\)-neighbourhood of \(x\in\mathbb{C}\) consists of all complex numbers being in the interior of a circle in the complex plane with midpoint \(x\) and radius \(\varepsilon\).
\([0,1]\) is a neighbourhood of \(\frac12\) (also of \(\frac34\), \(\frac1{\sqrt{2}}\) etc.), but it is not a neighbourhood of \(0\) or \(1\).
Definition 3 (Bounded, Open, closed, compact sets). Let \(M\subset \mathbb{K}\). Then \(M\) is called
bounded if there exists some \(c \in \mathbb{R}\) such that for all \(x\in M\) holds: \(|x|\leq c\).
open if for all \(x\in M\) holds: \(M\) is a neighbourhood of \(x\).
closed if for all convergent sequences \((a_n)_{n\in\mathbb{N}}\) with \(a_n\in M\) for all \(n\in\mathbb{N}\) holds: \(\lim_{n\to\infty}a_n=a\in M\).
compact if for all sequences \((a_n)_{n\in\mathbb{N}}\) with \(a_n\in M\) for all \(n\in\mathbb{N}\) holds: There exists some convergent subsequence \((a_{n_k})_{k\in\mathbb{N}}\) with \(\lim_{k\to\infty}a_{n_k}=a\in M\).
Example 4.
The interval \((0,1)\) is open.
Proof: Consider \(x\in(0,1)\). Then for \(\varepsilon=\min\{x,1-x\}\) holds \(\varepsilon>0\) and \(B_\varepsilon(x)=(x-\varepsilon,x+\varepsilon)\;\subset\;(0,1)\).\(\Box\)The interval \((0,1)\) is not closed.
Proof: Consider the sequence \((a_n)_{n\in\mathbb{N}}=(\frac1{n+1})_{n\in\mathbb{N}}\). Clearly, for all \(n\in\mathbb{N}\) holds \(a_n=\frac1{n+1}\in(0,1)\), but \((a_n)_{n\in\mathbb{N}}\) converges to \(0\notin(0,1)\). \(\Box\)The interval \((0,1)\) is not compact.
Proof: Again consider the sequence \((a_n)_{n\in\mathbb{N}}=(\frac1{n+1})_{n\in\mathbb{N}}\) in \((0,1)\). The convergence of \((a_n)_{n\in\mathbb{N}}\) to \(0\notin(0,1)\) also implies that this holds true for any subsequence \((a_{n_k})_{k\in\mathbb{N}}\) (see Theorem [thm:convsubseq]). Hence, any subsequence of the above constructed one is not convergent to some value in \((0,1)\). \(\Box\)The interval \((0,1]\) is neither open nor closed.
Proof: The closedness can be disproved by considering again the sequence \((a_n)_{n\in\mathbb{N}}=(\frac1{n+1})_{n\in\mathbb{N}}\), whereas the non-openness follows from the fact that \((0,1]\) is not a neighbourhood of \(1\). \(\Box\)The set \(\mathbb{R}\) is open and closed but not compact.
Proof: Openness and closedness are easy to verify. To see that this set is not compact, consider the sequence \((a_n)_{n\in\mathbb{N}}=(n)_{n\in\mathbb{N}}\) (which is of course in \(\mathbb{R}\)). It can be readily verified that any subsequence \((a_{n_k})_{k\in\mathbb{N}}=(n_k)_{k\in\mathbb{N}}\) is unbounded, too. Therefore, arbitrary subsequences \((a_{n_k})_{k\in\mathbb{N}}=(n_k)_{k\in\mathbb{N}}\) cannot converge.\(\Box\)The empty set \(\emptyset\) is open, closed and compact.
Proof: \(\emptyset\) is a neighbourhood of all \(x\in\emptyset\) (there is none, but the statement “for all \(x\in\emptyset\)” holds then true more than ever). By the same kind of argumentation, we can show that this set is compact and closed. The non-existence of a sequence in \(\emptyset\) implies that every statement holds true for them. In particular, all sequences \((a_{n})_{n\in\mathbb{N}}\) in \(\emptyset\) converge to some \(x\in\emptyset\) and have a convergent subsequence with limit in \(\emptyset\).\(\Box\)
Next we relate these three concepts to each other.
Theorem 5. For a set \(C\subset \mathbb{K}\), the following statements are equivalent:
\(C\) is open;
\(\mathbb{K}\backslash C\) is closed.
Proof:
“(i)\(\Rightarrow\)(ii)”: Let \(C\) be open. Consider a convergent sequence
\((a_{n})_{n\in\mathbb{N}}\) with \(a_n\in \mathbb{K}\backslash C\). We have to
show that for \(a=\lim_{n\to\infty}a_n\) holds \(a\in \mathbb{K}\backslash C\). Assume the
converse, i.e., \(a\in C\). Since \(C\) is open, we have that \(B_{\varepsilon}(a)\subset C\) for some
\(\varepsilon>0\). By the definition
of convergence, there exists some \(N\)
such that for all \(n\geq N\) holds
\(|a-a_n|<\varepsilon\), i.e., \[a_n\in B_{\varepsilon}(a)\subset C.\]
However, this is a contradiction to \(a_n\in
\mathbb{K}\backslash C\).
“(ii)\(\Rightarrow\)(i)”: Let \(\mathbb{K}\backslash C\) be closed. We have
to show that \(C\) is open. Assume the
converse, i.e., \(C\) is not open. In
particular, this means that there exists some \(a\in C\) such that for all \(n\in\mathbb{N}\) holds \(B_{\frac1n}(a)\not\subset C\). This means
that for all \(n\in \mathbb{N}\), we
can find some \(a_n\in \mathbb{K}\backslash
C\) with \(a_n\in
B_{\frac1n}(a)\), i.e., \(|a-a_n|<\frac1n\). As a consequence, for
the sequence \((a_n)_{n\in\mathbb{N}}\)
holds that \[\lim_{n\to\infty}a_n=a\in
C,\] but \(a_n\in \mathbb{K}\backslash
C\) for all \(n\in\mathbb{N}\).
This is a contradiction to the closedness of \(\mathbb{K}\backslash C\).\(\Box\)
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