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 compact implies every infinite subset \(K\) of E has a limit point in E, and which implies E is closed and bounded. 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. Then \(K\) must have a limit point in \(E\). If \(E\) is not bounded, then \(E\) contains points \(\mathbf x_n\) with \(\mathbf x_n>n, (n=1,2,3,\cdots)\). The set \(S\) consisting these points \(\mathbf x_n\) is infinite and has no limit point in \(R^k\), hence has none in \(E\), then “Every infinite subset \(K\) of E has a limit point in E” implies that \(E\) is bounded. If \(E\) is not closed, then there is a point \(\mathbf x_0\in R^k\) which is a limit point of \(E\) but not a point of \(E\). There is set \(S\) which contains the points \(\mathbf x_n\in E, n=1,2,3,\cdots\), such that \(|\mathbf x_n-\mathbf x_0|<1/n\), then \(S\) is infinite and \(S\) has \(\mathbf x_0\) as a limit point and \(S\) has no other limit point in \(R^k\). For if \(\mathbf y\in R^k, \mathbf y\ne \mathbf x_0\), then \[|\mathbf x_n-\mathbf y|\ge|\mathbf x_0-\mathbf y|-|\mathbf x_n-\mathbf x_0|\ge|\mathbf x_0-\mathbf y|-\frac{1}{n}\ge\frac{1}{2}|\mathbf x_0-\mathbf y|\] for all but finitely many \(n\). This shows that \(\mathbf y\) is not a limit point of \(S\). Thus \(S\) has no limit points in \(E\). Thus if “every infinite subset \(K\) of E has a limit point in E”, \(E\) must be closed.
A mapping \(f\) of a metric space \(X\) into a metric space \(Y\) is continuous on \(X\) if and only if \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). Suppose \(f\) is continuous on \(X\) and \(V\) is an open set in \(Y\), we have to show that every point of \(f^{-1}(V)\) is an interior point of \(f^{-1}(V)\). Suppose \(p\in X\) and \(f(p)\in V\), since \(V\) is open, there exists \(\varepsilon>0\) that \(y\in V\) if \(d_Y(f(p),y)<\varepsilon\) and since \(f\) is continuous at \(p\), there exists \(\delta>0\) such that \(d_Y(f(x),f(p))<\varepsilon\) if \(d_X(x,p)<\delta\). Thus \(x\in f^{-1}(V)\) if \(d_X(x,p)<\delta\). Conversely, suppose \(f^{-1}(V)\) is open in \(X\) for every open set \(V\) in \(Y\). Fix \(p\in X\) and \(\varepsilon>0\), let \(V\) be the set of all \(y\in Y\) such that \(d_Y(y,f(p))<\varepsilon\), then \(V\) is open. Hence \(f^{-1}(V)\) is open, there exists \(\delta>0\) such that \(x\in f^{-1}(V)\) if \(d_X(p,x)<\delta\). But if \(x\in f^{-1}(V)\), then \(f(x)\in V\), so that \(d_Y(f(x),f(p))<\varepsilon\), then \(f\) is continuous on \(X\).
Suppose \(f\) is a continuous mapping of a compact metric space \(X\) into a metric space \(Y\), then \(f(X)\) is compact. Let \(\{V_a\}\) be an open cover of \(f(X)\), since \(f\) is continuous so each of the sets \(f^{-1}(V_a)\) is open. Since \(X\) is compact, there are finitely many indices \(a_1,a_2,\cdots,a_n\), such that \[X\subset f^{-1}(V_{a_1})\cup\cdots\cup f^{-1}(V_{a_n})\] Since \(f(f^{-1}(E))\subset E\) for every \(E\subset Y\), then \[f(X)\subset V_{a_1}\cup \cdots\cup V_{a_n}\]
If \(\mathbf f\) is a continuous mapping of a compact metric space \(X\) into \(R^k\), then \(\mathbf f(X)\) is closed and bounded. Thus \(\mathbf f\) is bounded. When \(f\) is continuous real function on a compact metric space \(X\), then there exist points \(p,q\in X\) that \(f(p)=\text{sup } f(X)\) and \(f(q)=\text{inf } f(X)\).
Let \(f\) be a mapping of a metric space \(X\) into a metric space \(Y\), we say \(f\) is uniformly continuous on \(X\) if every \(\epsilon>0\) there exists \(\delta>0\) such that \(d_Y(f(p),f(q))<\epsilon\) for all \(p\) and \(q\) in \(X\) for which \(d_X(p,q)<\delta\).
Let \(f\) be a continuous mapping of a compact metric space \(X\) into a metric space \(Y\), then \(f\) is uniformly continuous on \(X\). Given \(\epsilon>0\), since \(f\) is continuous, we can associate each point \(p\in X\) a positive number \(\phi(p)\) such that \(\underset{q\in X}{d_X(p,q)}<\phi(p)\) implies that \(d_Y(f(p), f(q))<\frac{\epsilon}{2}\). Let \(J(p)\) be the set of all \(q\in X\) for which \[d_X(p,q)<\frac{1}{2}\phi(p)\] since \(p\in J(p)\), the collection of all sets \(J(p)\) is an open cover of \(X\); and since \(X\) is compact, there is a finite set of points \(p_1,\cdots,p_n\) in \(X\) that \(X\subset J(p_1)\cup\cdots\cup J(p_n)\). Let \(\delta=\frac{1}{2}\text{min}[\phi(p_1),\cdots,\phi(p_n)]\), then \(\delta>0\). Now let \(p,q\) be the points of \(X\), such that \(d_X(p,q)<\delta\), since \(X\subset J(p_1)\cup\cdots\cup J(p_n)\), there is an integer \(m, 1\le m\le n\) such that \(p\in J(p_m)\), hence \(d_X(p,p_m)<\frac{1}{2}\phi(p_m)\) and we also have \[d_X(q,p_m)\le d_X(q,p)+d_X(p,p_m)<\delta+\frac{1}{2}\phi(p_m)\le \phi(p_m)\] therefore \[d_Y(f(p),f(q))\le d_Y(f(p),f(p_m))+d_Y(f(q),f(p_m))<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\] which means \(f\) is uniformly continuous on \(X\).
Let \(E\) be a noncompact set in \(R^1\), then there exists a continuous mapping on \(E\) which is unbounded. (a) Suppose \(E\) is bounded, because \(E\) is noncompact, there exists a limit point \(x_0\) of \(E\) which is not a point of \(E\). Consider \[f(x)=\frac{1}{x-x_0}\quad(x\in E)\] This is continuous on \(E\) and unbounded. To see this function is not uniformly continuous, let \(\epsilon>0\) and \(\delta>0\), and choose a point \(x\in E\) such that \(|x-x_0|<\delta\). Taking \(t\) close enough to \(x_0\), we can then make \(|f(t)-f(x)|\) greater than \(\epsilon\), although \(|t-x|<\delta\).Since this is true for every \(\delta>0\), \(f\) is not uniformly continuous on \(E\). (b) Suppose \(E\) is unbounded, such that \(f(x)=x\), which is continuous on \(E\) and unbounded.
Let \(E\) be a noncompact set in \(R^1\), then there exists a continuous and bounded mapping on \(E\) which has no maximum. The given function \[g(x)=\frac{1}{1+(x-x_0)^2},\quad(x\in E)\] is continuous on \(E\) and is bounded, since \(<0g(x)<1\). It is clear that \[\underset{x\in E}{\text{sup }}g(x)=1\] whereas \(g(x)<1\) for all \(x\in E\). Thus \(g(x)\) has no maximum on \(E\).
Let \(f\) be defined on \((a,b)\). If \(f\) is discontinuous at the point \(x\), and if \(f(x+)\) and \(f(x-)\) exist, but \(f(x+)\ne f(x-)\) or \(f(x+)=f(x-)\ne f(x)\), then \(f\) is said to have first kind discontinuity or simple discontinuity at \(x\). Otherwise, if \(f(x+)\) or \(f(x-)\) does not exist, the discontinuity is called second kind discontinuity.