Let V = C (1) ([0, 1]), || f || . Find the interior, closure,Isolated points, and boundary of M = {ce x |c R}. Here C (1) ([0, 1] is the space of continuously differentiable functions on [0, 1], and || f || = supx[a,b] | f (x)|. 1 \\documentclass[12pt]{article} \\usepackage[all]{xy} \\begin{document} Interior points: Let $\\epsilon$ >o , then $ce^x + \\epsilon$ does not belongs to $M$.But the function $ce^x + \\epsilon$ is a continously differential function. Hence any ball containing any member of $M$ does not contained in $M$. Hence $int(M) = \\phi$. Closure : Let $f \\in C^{(1)}([0,1])$ such that $f \\in closure(M)$. then $\\exists \\{ c_ne^x\\} \\subset M$ Such that $f(x)= lim_{n \ ightarrow \\infty}c_ne^x = e^xlim_{n \ ightarrow \\infty}c_n$ This imply $f(x)= ce^x$ where $c =lim_{n \ ightarrow \\infty}c_n$ $\\Rightarrow$ $f \\in M$ $ \\Rightarrow $ Mis a closed set.Hence $closure(M)=M.$ Isolated points: Let for some $c_{0}$ , $c_oe^x \\in M$ is an isolated point.Then there is $\\delta$>0 ,such that $M \\cap B_{\\delta}(c_0 e^x) = \\{c_0e^x \\}$. i.e. $||f - c_0e^x|| < \\delta$ for all $f$ in the ball .Now choose $f = ((\\delta / ||e^x||_{\\infty}) + c_0)e^x$ then $f \\in M \\cap B_{\\delta}(c_0 e^x)$ this implies there is no isolated points. Boundary : $\\delta(M) = closure(M) -int(M)= M - \\phi = M$. Hence boundary of $M (= \\delta(M))$ is full of $M$ \\end{document} \\documentclass[12pt]{article} \\usepackage[all]{xy} \\begin{document} Interior points: Let $\\epsilon$ >o , then $ce^x + \\epsilon$ does not belongs to $M$.But the function $ce^x + \\epsilon$ is a continously differential function. Hence any ball containing any member of $M$ does not contained in $M$. Hence $int(M) = \\phi$. Closure : Let $f \\in C^{(1)}([0,1])$ such that $f \\in closure(M)$. then $\\exists \\{ c_ne^x\\} \\subset M$ Such that $f(x)= lim_{n \ ightarrow \\infty}c_ne^x = e^xlim_{n \ ightarrow \\infty}c_n$ This imply $f(x)= ce^x$ where $c =lim_{n \ ightarrow \\infty}c_n$ $\\Rightarrow$ $f \\in M$ $ \\Rightarrow $ Mis a closed set.Hence $closure(M)=M.$ Isolated points: Let for some $c_{0}$ , $c_oe^x \\in M$ is an isolated point.Then there is $\\delta$>0 ,such that $M \\cap B_{\\delta}(c_0 e^x) = \\{c_0e^x \\}$. i.e. $||f - c_0e^x|| < \\delta$ for all $f$ in the ball .Now choose $f = ((\\delta / ||e^x||_{\\infty}) + c_0)e^x$ then $f \\in M \\cap B_{\\delta}(c_0 e^x)$ this implies there is no isolated points. Boundary : $\\delta(M) = closure(M) -int(M)= M - \\phi = M$. Hence boundary of $M (= \\delta(M))$ is full of $M$ \\end{document}