In metric space, a neighborhood of \(p\) is a set \(N_r(p)\) consisting of all \(q\) such that \(d(p,q)<r\), for some \(r>0\). The \(r\) is called radius of \(N_r(p)\). A point \(p\) is called a limit point of the set \(E\), if every neighborhood of \(p\) contains a point \(q\ne p\) and \(q\in E\). \(E\) is called closed if every limit point of \(E\) is a point of \(E\). \(E\) is called perfect if \(E\) is closed and every point of \(E\) is a limit point of \(E\). \(p\) is called an interior point of \(E\) if there is a neighborhood \(N_r(p)\) of \(p\) such that \(N_r(p)\subset E\). \(E\) is open if every point of \(E\) is an interior point of \(E\). Every neighborhood is an open set because every point in the neighborhood set \(N\) is an interior point of \(N\). Subset \(E\) of metric space \(X\) is called bounded if for all points \(p\in E\), \(d(p,q)<M\) for a real number \(M\) and a point \(q\in X\).
Suppose \(Y\subset X\) and \(E\) is open relative to \(Y\), then, to each \(p\in E\), there is a positive number \(r_p\) and the condition \(q\in Y\), \(d(p,q)<r_p\) imply that \(q\in E\). Let \(V_p\) be the set of all \(q\in X\) such that \(d(p,q)<r_p\) and define \[G=\underset{p\in E}{\cup} V_p\] Then \(G\) is an open subset of \(X\), since \(p\in V_p\) for all \(p\in E\), it is clear that \(E\subset G\cap Y\). By our choice of \(V_p\), we have \(V_p\cap Y\subset E\) for every \(p\in E\), so that \(G\cap Y\subset E\). Then \[E=G\cap Y\] Conversely, if \(G\) is open in \(X\) and \(E=G\cap Y\), every \(p\in E\) has a neighborhood \(V_p\in G\), then \(V_p\cap Y\subset E\), so that \(E\) is open relative to \(Y\). Then, suppose \(Y\subset X\), a subset \(E\) of \(Y\) is open relative to \(Y\) if and only if \(E=Y\cap G\) for some open subset \(G\) of \(X\).
The collection of open subsets \(\{G_{\alpha}\}\) of a metric space \(X\) such that \(E\subset\underset{\text{all }\alpha}{\bigcup} G_{\alpha}\) is called an open cover of set \(E\). For a subset \(K\) of a metric space \(X\), if every open cover of \(K\) contains a finite subcovers, then subset \(K\) is said to be compact. Explicitly, if the collection of open subsets \(\underset{\text{all }\alpha}{\bigcup} G_{\alpha}\) is an open cover of \(K\), then there are finitely many indices \(\alpha_1,\alpha_2,\cdots,\alpha_n\) such that \[K\subset G_{\alpha_1}\cup G_{\alpha_2}\cup\cdots\cup G_{\alpha_n}\]
Suppose \(K\subset Y\subset X\) and \(K\) is compact relative to \(X\), let \(\{V_{\alpha}\}\) be a collection of sets and open relative to \(Y\), and \(K\subset\underset{\text{all }\alpha}{\cup} V_{\alpha}\), then there are sets \(G_{\alpha}\) open relative to \(X\) with \(V_{\alpha}=Y\cap G_{\alpha}\) for all \(\alpha\); and since \(K\) is compact relative to \(X\), we have \[K\subset G_{\alpha_1}\cup G_{\alpha_2}\cup\cdots\cup G_{\alpha_n}\] for some choice of finitely many indices \(\alpha_1, \cdots, \alpha_n\). Since \(K\subset Y\), then \[K\subset Y\cap(G_{\alpha_1}\cup G_{\alpha_2}\cup\cdots\cup G_{\alpha_n})\Rightarrow K\subset V_{\alpha_1}\cup\cdots\cup V_{\alpha_n}\] this means \(K\) is compact relative to \(Y\). Conversely, suppose \(K\) is compact relative to \(Y\), let \(G_{\alpha}\) be a collection of open subsets of \(X\) which covers \(K\), and \(V_{\alpha}=Y\cap G_{\alpha}\) Then \[K\subset V_{\alpha_1}\cup\cdots\cup V_{\alpha_n}\] will hold for some choice of finitely many indices \(\alpha_1, \cdots, \alpha_n\) and since \(V_{\alpha}\subset G_{\alpha}\), then \(K\subset G_{\alpha_1}\cup G_{\alpha_2}\cup\cdots\cup G_{\alpha_n}\) and then \(K\) is compact relative to \(X\).
Suppose \(E^c\) is closed, \(x\in E\), then \(x\notin E^c\) and \(x\) is not a limit point of \(E^c\). Hence there exists a neighborhood \(N\) of \(x\) such that \(E^c\cap N\) is empty, that is \(N\in E\). Thus \(p\) is an interior point of \(E\) and \(E\) is open. Next suppose \(E\) is open, let \(x\) be a limit point of \(E^c\). Then every neighborhood of \(x\) contains a point of \(E^c\), so that \(x\) is not an interior point of \(E\), since \(E\) is open, this means that \(x\in E^c\) and \(E^c\) is closed. The conclusion is a set \(E\) is open if and only if its complement is closed or a set \(E\) is closed if and only if its complement is open.
Let \(K\) be a compact subset of metric space \(X\), suppose \(q\in K\), \(p\in X\) and \(p\notin K\), let \(V_q\) and \(W_q\) be the neighborhoods of \(p\) and \(q\), respectively, of radius less than \(\frac{1}{2}d(p,q)\). Since \(K\) is compact there are finitely many points \(q_1,\cdots,q_n\) in \(K\), such that \[K\subset (W_{q1}\cup\cdots\cup W_{qn})=W\] if \(V=V_{q1}\cap\cdots\cap V_{qn}\), then \(V\) is a neighborhood of \(p\) which does not intersect with \(W\). Then \(V\subset K^c\), so that \(p\) is an interior point of \(K^c\), which means \(K^c\) is open and \(K\) is closed.
Closed subsets of a compact set are compact. Suppose the closed subset of compact set \(F\subset K\subset X\), \(F\) is a closed subset relative to compact set \(K\), let \(\{V_{\alpha}\}\) be an open cover of \(F\), if \(F^c\) is adjoined to \(\{V_{\alpha}\}\), we obtain an open cover \(\Omega\) of \(K\). Since \(K\) is compact, there is finite sub-collection \(\Phi\) of \(\Omega\) which covers \(K\), and hence \(F\). If \(F^c\) is a member of \(\Phi\), we may remove it from \(\Phi\) and still retain an open cover of \(F\). Then, a finite sub-collection of \(\{V_{\alpha}\}\) covers \(F\) and \(F\) is compact.
If \(\{K_{\alpha}\}\) is a collection of compact subsets of a metric space \(X\) such that the intersection of every finite sub-collection of \(\{K_{\alpha}\}\) is nonempty, fix a member \(K_1\) of \(\{K_{\alpha}\}\) and denote \(G_{\alpha}=K_{\alpha}^c\). Assume that no point of \(K_1\) belongs to every \(K_{\alpha}\), then the sets \(G_{\alpha}\) form an open cover of \(K_1\) and since \(K_1\) is compact, there are finitely many indices \(\alpha_1,\cdots,\alpha_n\), such that \[K_1\subset(G_{\alpha_1}\cup\cdots\cup G_{\alpha_n})\] or \[K_1\subset(K_{\alpha_1}^c\cup\cdots\cup K_{\alpha_n}^c)\] or \[K_1\subset(K_{\alpha_1}\cap\cdots\cap K_{\alpha_n})^c\] but this means \[K_1\cap(K_{\alpha_1}\cap\cdots\cap K_{\alpha_n})\] is empty, which contradicts to our hypothesis. Then there must be some point of \(K_1\) belongs to every \(K_{\alpha}\) and \(\cap K_{\alpha}\) is nonempty.
If \(E\) is an infinite subset of compact set \(K\), then \(E\) must have a limit point in \(K\). Because if no point of \(K\) is a limit point of \(E\), each point \(q\in K\) would have a neighborhood \(V_q\) which contains at most one point of \(E\) (if there are more than one points in \(E\) for every neighborhood of \(q\), then \(q\) will be a limit point of \(E\)). And because \(E\) is infinite, it is clear that no finite sub-collection of \(\{V_q\}\) can cover \(E\) and the same is true of \(K\), since \(E\subset K\), then \(K\) can’t be compact.
The set of all real numbers \(x\) such that \(a<x<b\) is \((a,b)\), which is called segment. The set of all real numbers \(x\) such that \(a\le x\le b\) is \([a,b]\), which is called interval. If points of \(k\)-dimension \[\mathbf a=\begin{bmatrix} a_1\\ a_2\\ \vdots\\ a_k\\ \end{bmatrix}<\mathbf b=\begin{bmatrix} b_1\\ b_2\\ \vdots\\ b_k\\ \end{bmatrix}\] means \(a_i<b_i,i=1,2,\cdots,k\), the sets of all points \(\mathbf x\) satisfy \(\mathbf a\le\mathbf x\le\mathbf b\) is called a k-cell. Thus the interval is a 1-cell.
Every k-cell is compact, because let \(\delta=\Biggl[\displaystyle\sum_{i=1}^{k}(a_i-b_i)^2\Biggr]^{1/2}\) be the length of the diagonal of this k-cell \(I\), then for any two points \(\mathbf x\) and \(\mathbf y\) in \(I\), \(|\mathbf x-\mathbf y|\le\delta\). Let \(c_i=(a_i+b_i)/2\), then \(I\) can be divided to \(2^k\) k-cells by \([a_i,c_i]\) and \([c_i,b_i]\), the diagonal of this k-cell \(I_1\) is \(2^{-1}\delta\). We can divide \(I_1\) to \(2^k\) k-cells again to get \(I_2\), the diagonal of these new cells is \(2^{-2}\delta\), then the diagonal of \(I_n\) is \(2^{-n}\delta\). If the radius of open cover \(G_{\alpha}\) is \(r\), we can increase \(n\) large enough that the diagonal of the k-cell \(I_n\) \(2^{-n}\delta<r\) then \(I_n\subset G_{\alpha}\) which means every k-cell can be covered by some sub-collection of \(\{G_{\alpha}\}\) and the number of these sub-collections \(\{G_{\alpha}\}\) is finite and no more than \(2^{kn}\).
If a set \(E\) in \(R^k\) is closed and bounded, then \(E\subset I\) for some compact k-cell \(I\), then \(E\) is the closed subset of compact set \(I\) and \(E\) is also compact. A bounded infinite set \(E\) in \(R^k\) is a subset of a compact k-cell \(I\), and \(E\) must have a limit point in \(I\) or \(R^k\).
\(E\) is called perfect if \(E\) is closed and every point of \(E\) is a limit point of \(E\). If \(E\) is a nonempty perfect set in \(R^k\), then \(E\) is uncountable. Since \(E\) is perfect, \(E\) contains limit points and must be infinite. If \(E\) is countable such as \(\mathbf x_1,\mathbf x_2,\mathbf x_3,\cdots\), we can construct a sequence of neighborhoods \(\{V_n\}\) as follows: \(V_1\) is the neighborhood of \(\mathbf x_1\) and consists of all points \(\mathbf y\in R^k\) that \(|\mathbf y-\mathbf x_1|<r\), and the closure of \(V_1\) is \(\bar{V_1}\) which is the set of all \(\mathbf y\) that \(|\mathbf y-\mathbf x_1|\le r\). Then \(V_n\cap E\) is nonempty because every point of \(E\) is a limit point, there is a neighborhood \(V_{n+1}\), such that \(\bar{V}_{n+1}\subset V_n\), \(\mathbf x_n\notin V_{n+1}\) and \(\bar{V}_{n+1}\cap E\) is nonempty. Denote \(K_n=\bar{V}_{n+1}\cap E\), since \(\bar{V}_{n+1}\) is closed and bounded, \(\bar{V}_{n+1}\) is compact. Since \(\mathbf x_n\notin V_{n+1}\), no points of \(E\) lies in \(\cap_{n=1}^{\infty} K_n\), this implies \(\cap_{n=1}^{\infty} K_n\) is empty, but since \(K_{n+1}\subset K_n\) and each \(K_n\) is nonempty, this contradiction means \(E\) is uncountable.
\(X\) is a Hausdorff space if the following is true: If \(p\in X,q\in X\) and \(p\ne q\), then \(p\) has a neighborhood \(U\) and \(q\) has a neighborhood \(V\) such that \(U\cap V=\varnothing\). \(X\) is locally compact if every point of \(X\) has a neighborhood whose closure is compact. every compact space is locally compact. Since The compact subsets of a euclidean space \(R^n\) are precisely those that are closed and bounded, then \(R^n\) is a locally compact Hausdorff space, and every metric space is a Hausdorff space.
Suppose \(X\) is a Hausdorff space, \(K\subset X\), \(K\) is compact, and \(p\in K^c\). Then there are open sets \(U\) and \(W\) such that \(p\in U, K\subset W\) and \(U\cap W=\varnothing\).
If \(q\in K\), there exist disjoint open sets \(U_q\) and \(V_q\), such that \(p\in U_q\) and \(q\in V_q\). Since \(K\) is compact, there are points \(q_1,q_2,\cdots,q_n\in K\) such that \[K\subset V_{q_1}\cup\cdots\cup V_{q_n}\] Then \[U=U_{q_1}\cap\cdots\cap U_{q_n}\] and \[W=V_{q_1}\cup\cdots\cup V_{q_n}\] satisfy that \(p\in U, K\subset W\) and \(U\cap W=\varnothing\). Since \(p\) can be every point in \(K^c\), then \(K^c\) must be open and \(K\) must be closed. Compact subsets of Hausdorff spaces are closed.If \(\{K_a\}\) is a collection of compact subsets of a Hausdorff space and if \(\bigcap_aK_a=\varnothing\). Fix a member \(K_1\) of \(\{K_a\}\), since no point of \(K_1\) belongs to every other \(K_a\), \(\{K_a^c\}\) is an open cover of \(K_1\). Hence \(K_1\subset K_{a_1}^c\cup\cdots\cup K_{a_n}^c\) for some finite collection \(\{K_{a_i}^c\}\). This implies that \[K_1\cap K_{a_1}\cap\cdots\cap K_{a_n}=\varnothing\] then some finite sub-collection of \(\{K_a\}\) also has empty intersection.
There is open set with compact closure between open set and its compact subset Suppose \(U\) is open in a locally compact Hausdorff space \(X\), \(K\subset U\), and \(K\) is compact. Then there is an open set \(V\) with compact closure such that \[K\subset V\subset \overline{V}\subset U\]
Since every point of \(K\) has a neighborhood with compact closure, and since \(K\) is covered by the union of finitely many of these neighborhoods, \(K\) lies in an open set \(G\) with compact closure. If \(U=X\), take \(V=G\). Otherwise, the theorem in 15) shows that, to each \(p\in U^c\) there corresponds an open set \(W_p\) such that \(K\subset W_p\) and \(p\notin\overline{W}_p\). Hence \(\{U^c\cap\overline{G}\cap\overline{W}_p\}\), where \(p\) ranges over \(U^c\), is a collection of compact sets with empty intersection. The theorem in 16) shows that there are points \(p_1,\cdots,p_n\in U^c\) such that \[U^c\cap\overline{G}\cap\overline{W}_{p_1}\cap\cdots\cap\overline{W}_{p_n}=\varnothing\] The set \[V=G\cap W_{p_1}\cap\cdots\cap W_{p_n}\] with compact closure \(\overline{V}\) satisfy that \[K\subset V\subset \overline{V}\subset U\] since \[\overline{V}\subset \overline{G}\cap \overline{W}_{p_1}\cap\cdots\cap \overline{W}_{p_n}\]Let \(f\) be a real function on a topological space. If \[\{x:f(x)>a\}\] is open for every real \(a\), \(f\) is said to be lower semicontinuous. If \[\{x:f(x)<a\}\] is open for every real \(a\), \(f\) is said to be upper semicontinuous. The supremum of any collection of lower semicontinuous functions is lower semicontinuous. The infimum of any collection of upper semicontinuous functions is upper semicontinuous.
The support of a complex function \(f\) on a topological space \(X\) is the closure of the set \[\{x:f(x)\ne0\}\] The collection of all continuous complex functions on \(X\) whose support is compact is denoted by \(C_c(X)\), which is a vector space.
continuous mapping is compact domain to compact range, or the range of any \(C_c(X)\) is compact. Let \(X\) and \(Y\) be topological spaces, and let \(f:X\to Y\) be continuous. If \(K\) is a compact subset of \(X\), then \(f(K)\) is compact.
If \(\{V_a\}\) is an open cover of \(f(K)\), then \(\{f^{-1}(V_a)\}\) is an open cover of \(K\), hence \(K\subset f^{-1}(V_{a_1})\cup\cdots\cup f^{-1}(V_{a_n})\) for some \(a_1,\cdots,a_n\) and therefore \(f(K)\subset V_{a_1}\cup\cdots\cup V_{a_n}\)The notation \(K\prec f\) means that \(K\) is a compact subset of \(X\), that \(f\in C_c(X),\;\;0\leq f(x)\leq 1\) for all \(x\in X\) and that \(f(x)=1\) for all \(x\in K\). The notation \[f\prec V\] means that \(V\) is open, that \(f\in C_c(X), \; 0\leq f(x)\leq 1\), and that the support of \(f\) lies in \(V\).
References:
1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.