IGNOU MMT-004 Solved Assignment 2024 | M.Sc. MACS Abstract Classes ®

IGNOU MMT-004 Solved Assignment 2024 for M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMT-004 Assignment Question Paper 2024

mmt-003-solved-assignment-2024-qp-2afdf4f2-c576-4553-8e87-8dfd8225b98e

MMT-003 Solved Assignment 2024 QP

a) Let $X = C [0, 1]$ $X = C [0, 1]$ X=C[0,1]mathrm{X}=mathrm{C}[0,1] $X = C [0, 1]$ . Define $d : X \times X \to R$ $d : X \times X \to R$ d:X xx X rarrRd: X times X rightarrow mathbf{R} $d : X \times X \to R$ by $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginXmathrm{d}(mathrm{f}, mathrm{g})=int_0^1|mathrm{f}(mathrm{t})-mathrm{g}(mathrm{t})| mathrm{dt}, mathrm{f}, mathrm{g} in mathrm{X} $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ where the integral is the Riemann integral. Show that $d$ $d$ dd $d$ is a metric on $X$ $X$ XX $X$ . Find $d (f, g)$ $d (f, g)$ d(f,g)d(f, g) $d (f, g)$ where $f (x) = 4 x$ $f (x) = 4 x$ f(x)=4xf(x)=4 x $f (x) = 4 x$ and $g (x) = x^{3}, x \in [0, 1]$ $g (x) = x^{3}, x \in [0, 1]$ g(x)=x^(3),x in[0,1]g(x)=x^3, x in[0,1] $g (x) = x^{3}, x \in [0, 1]$ .

b) Let ( $X, d)$ $X, d)$ X,d)X, d) $X, d)$ be a metric space and $a \in X$ $a \in X$ a in Xa in X $a \in X$ be a fixed point of $X$ $X$ XX $X$ . Show that the function $f_{a} : X \to R$ $f_{a} : X \to R$ f_(a):X rarrRf_a: X rightarrow mathbf{R} $f_{a} : X \to R$ given by $f_{a} (x) = d (x, a)$ $f_{a} (x) = d (x, a)$ f_(a)(x)=d(x,a)mathrm{f}_{mathrm{a}}(mathrm{x})=mathrm{d}(mathrm{x}, mathrm{a}) $f_{a} (x) = d (x, a)$ is continuous. Is it uniformly continuous? Justify you answer.
2. a) Let $A$ $A$ AA $A$ and $B$ $B$ BB $B$ be any two subsets of a metric space (X, d), then show that
i) int $A = \cup {E$ $A = \cup {E$ A=uu{Emathrm{A}=cup{mathrm{E} $A = \cup {E$ : is open and $E \subseteq A}$ $E \subseteq A}$ EsubeA}mathrm{E} subseteq mathrm{A}} $E \subseteq A}$
ii) $int (A \cap B) = int A \cap int B$ $int (A \cap B) = int A \cap int B$ int(AnnB)=int Ann int Boperatorname{int}(mathrm{A} cap mathrm{B})=operatorname{int} mathrm{A} cap operatorname{int} mathrm{B} $int (A \cap B) = int A \cap int B$
iii) $int (A \cup B) \supseteq$ $int (A \cup B) \supseteq$ quad int(A uu B)supequad operatorname{int}(A cup B) supseteq $int (A \cup B) \supseteq$ int $A \cap$ $A \cap$ A nnA cap $A \cap$ int $B$ $B$ BB $B$
iv) $\bar{A \cap B} \subseteq \bar{A} \cap \bar{B}$ $\bar{A \cap B} \subseteq \bar{A} \cap \bar{B}$ bar(AnnB)sube bar(A)nn bar(B)overline{mathrm{A} cap mathrm{B}} subseteq overline{mathrm{A}} cap overline{mathrm{B}} $\bar{A \cap B} \subseteq \bar{A} \cap \bar{B}$ .
b) Find the interior, boundary and closure of the following sets $A$ $A$ Amathbf{A} $A$ in $R$ $R$ Rmathbf{R} $R$ with the usual metric and discrete metric.
i) $A = Q$ $A = Q$ A=QA=mathbf{Q} $A = Q$ , the set of rationals in $R$ $R$ Rmathbf{R} $R$
ii) $A =] 1, 2] \cup] 2, 4 [$ $A =] 1, 2] \cup] 2, 4 [$ A=]1,2]uu]2,4[mathrm{A}=] 1,2] cup] 2,4[ $A =] 1, 2] \cup] 2, 4 [$
3. a) Let $(X, d_{1})$ $(X, d_{1})$ (X,d_(1))left(X, d_1right) $(X, d_{1})$ and $(Y, d_{2})$ $(Y, d_{2})$ (Y,d_(2))left(Y, d_2right) $(Y, d_{2})$ be metric spaces. Show that $f : X \to Y$ $f : X \to Y$ f:X rarr Yf: X rightarrow Y $f : X \to Y$ is continuous if and only if $f (\bar{A}) \subseteq \bar{f (A)}$ $f (\bar{A}) \subseteq \bar{f (A)}$ f( bar(A))sube bar(f(A))f(bar{A}) subseteq overline{f(A)} $f (\bar{A}) \subseteq \bar{f (A)}$ where $A$ $A$ AA $A$ is any subset of $X$ $X$ XX $X$
b) Let $(X_{1}, d_{1})$ $(X_{1}, d_{1})$ (X_(1),d_(1))left(mathrm{X}_1, mathrm{~d}_1right) $(X_{1}, d_{1})$ and $(X_{2}, d_{2})$ $(X_{2}, d_{2})$ (X_(2),d_(2))left(mathrm{X}_2, mathrm{~d}_2right) $(X_{2}, d_{2})$ be two discrete metric spaces. Verify that the product metric on $X_{1} \times X_{2}$ $X_{1} \times X_{2}$ X_(1)xxX_(2)mathrm{X}_1 times mathrm{X}_2 $X_{1} \times X_{2}$ is discrete.
c) Show that an infinite discrete metric space $X$ $X$ Xmathrm{X} $X$ is bounded but not totally bounded.
4. a) Find the first derivative $f^{'} (a)$ $f^{'} (a)$ f^(‘)(a)mathrm{f}^{prime}(mathbf{a}) $f^{'} (a)$ of the function $f$ $f$ fmathrm{f} $f$ defined by $f : R^{3} \to R^{2}$ $f : R^{3} \to R^{2}$ f:R^(3)rarrR^(2)f: mathbf{R}^3 rightarrow mathbf{R}^2 $f : R^{3} \to R^{2}$ given by $f (x, y, z) = (x y z, x + y + z^{2})$ $f (x, y, z) = (x y z, x + y + z^{2})$ f(x,y,z)=(xyz,x+y+z^(2))f(x, y, z)=left(x y z, x+y+z^2right) $f (x, y, z) = (x y z, x + y + z^{2})$ where $a = (1 . - 1, 2)$ $a = (1 . - 1, 2)$ a=(1.-1,2)mathbf{a}=(1 .-1,2) $a = (1 . - 1, 2)$ .
b) Let $E$ $E$ Emathrm{E} $E$ be an open subset of $R^{n}$ $R^{n}$ R^(n)mathbf{R}^n $R^{n}$ and $f : E \to R^{m}$ $f : E \to R^{m}$ f:E rarrR^(m)f: E rightarrow mathbf{R}^m $f : E \to R^{m}$ be a function such that each of its components function $f_{i}$ $f_{i}$ f_(i)f_i $f_{i}$ are differentiable, then show that $f$ $f$ ff $f$ is differentiable. Is the converse of this result true? Justify your answer.
c) Near what points may the surface $z^{2} + x z + y = 0$ $z^{2} + x z + y = 0$ z^(2)+xz+y=0z^2+x z+y=0 $z^{2} + x z + y = 0$ be represented uniquely as a graph of a differentiable function $z = k (x, y)$ $z = k (x, y)$ z=k(x,y)mathrm{z}=mathrm{k}(mathrm{x}, mathrm{y}) $z = k (x, y)$ ? Locate such a point.
5. a) Use the method of Lagrange’s multiplier method to find the shortest possible distance from the ellipse $x^{2} + 2 y^{2} = 2$ $x^{2} + 2 y^{2} = 2$ x^(2)+2y^(2)=2x^2+2 y^2=2 $x^{2} + 2 y^{2} = 2$ to the line $x + y = 2$ $x + y = 2$ x+y=2x+y=2 $x + y = 2$ .
b) Find the directional derivative of the function $f : R^{4} \to R^{3}$ $f : R^{4} \to R^{3}$ f:R^(4)rarrR^(3)f: mathbf{R}^4 rightarrow mathbf{R}^3 $f : R^{4} \to R^{3}$ defined by
$f (x, y, z, w) = (x^{2} y, x y z, x^{2} + y^{2} + z^{2})$ $f (x, y, z, w) = (x^{2} y, x y z, x^{2} + y^{2} + z^{2})$ f(x,y,z,w)=(x^(2)y,xyz,x^(2)+y^(2)+z^(2))f(x, y, z, w)=left(x^2 y, x y z, x^2+y^2+z^2right) $f (x, y, z, w) = (x^{2} y, x y z, x^{2} + y^{2} + z^{2})$
at $a = (1, 2, - 1, - 2)$ $a = (1, 2, - 1, - 2)$ a=(1,2,-1,-2)mathrm{a}=(1,2,-1,-2) $a = (1, 2, - 1, - 2)$ in the direction $v = (0, 1, 2, - 2)$ $v = (0, 1, 2, - 2)$ v=(0,1,2,-2)mathrm{v}=(0,1,2,-2) $v = (0, 1, 2, - 2)$ .
6. a) Let A be a compact non-empty subset of a metric space (X, d) and let F be a closed subset of $X$ $X$ XX $X$ such that $A \cap F = ϕ$ $A \cap F = ϕ$ A nn F=phiA cap F=phi $A \cap F = ϕ$ , then show that $d (A, F) > 0$ $d (A, F) > 0$ d(A,F) > 0d(A, F)>0 $d (A, F) > 0$ where $d (A, F) = inf {d (a, b) : a \in A, b \in F}$ $d (A, F) = inf {d (a, b) : a \in A, b \in F}$ d(A,F)=i n f{d(a,b):a in A,b in F}d(A, F)=inf {d(a, b): a in A, b in F} $d (A, F) = inf {d (a, b) : a \in A, b \in F}$ .
b) Give an example of the following with justification
i) A vector-valued function $f : R^{3} \to R^{3}$ $f : R^{3} \to R^{3}$ f:R^(3)rarrR^(3)f: mathbf{R}^3 rightarrow mathbf{R}^3 $f : R^{3} \to R^{3}$ which is not differentiable at $(0, 0, 0)$ $(0, 0, 0)$ (0,0,0)(0,0,0) $(0, 0, 0)$ .
ii) A function which is Legesgue measurable on $R$ $R$ Rmathbf{R} $R$ .
c) Show that the components of a metric space is either identical or pairwise disjoint.
7. a) Let $Q$ $Q$ Qmathbf{Q} $Q$ be the set of rationals with the metric defined on $Q$ $Q$ Qmathbf{Q} $Q$ by $d : Q \times Q \to R$ $d : Q \times Q \to R$ d:QxxQrarrRd: mathbf{Q} times mathbf{Q} rightarrow mathbf{R} $d : Q \times Q \to R$ , defined by $d (x, y) = | x - y |, \forall x, y \in R$ $d (x, y) = | x - y |, \forall x, y \in R$ d(x,y)=|x-y|,AA x,y inRd(x, y)=|x-y|, forall x, y in mathbf{R} $d (x, y) = | x - y |, \forall x, y \in R$ .
Show that ${{(1 + \frac{1}{n})}^{n}}$ $\{{(1 + \frac{1}{n})}^{n}\}$ {(1+(1)/(n))^(n)}left{left(1+frac{1}{mathrm{n}}right)^{mathrm{n}}right} ${{(1 + \frac{1}{n})}^{n}}$ is Cauchy sequence in $Q$ $Q$ Qmathbf{Q} $Q$ , but does not converge in $Q$ $Q$ Qmathbf{Q} $Q$ and ${\frac{1}{3^{n}}}$ $\{\frac{1}{3^{n}}\}$ {(1)/(3^(n))}left{frac{1}{3^n}right} ${\frac{1}{3^{n}}}$ is a Cauchy sequence $Q$ $Q$ Qmathbf{Q} $Q$ which converges in $Q$ $Q$ Qmathbf{Q} $Q$ to the limit 0 .
b) Which of the following sets are totally bounded? Give reasons for your answer. Are they compact?
i) $2 N$ $2 N$ quad2Nquad 2 mathbf{N} $2 N$ in $(N, d)$ $(N, d)$ (N,d)(mathbf{N}, d) $(N, d)$ where $d$ $d$ dd $d$ is the discrete metric.
ii) $[0, 2] \cup [5, 10]$ $[0, 2] \cup [5, 10]$ quad[0,2]uu[5,10]quad[0,2] cup[5,10] $[0, 2] \cup [5, 10]$ in $(R, d)$ $(R, d)$ (R,d)(mathbf{R}, d) $(R, d)$ where $d$ $d$ dd $d$ is the Euclidean metric.
c) Which of the following sets are connected sets in $R^{2}$ $R^{2}$ R^(2)mathbf{R}^2 $R^{2}$ with the metric given against it? Justify your answer.
i) $A = {(x, y) : 0 \leq x \leq 1, 0 \leq y \leq 2}$ $A = {(x, y) : 0 \leq x \leq 1, 0 \leq y \leq 2}$ quadA={(x,y):0 <= x <= 1,0 <= y <= 2}quad mathrm{A}={(mathrm{x}, mathrm{y}): 0 leq mathrm{x} leq 1,0 leq mathrm{y} leq 2} $A = {(x, y) : 0 \leq x \leq 1, 0 \leq y \leq 2}$ under the standard metric.
ii) $A = {(x, y) : x^{2} + y^{2} = 1}$ $A = \{(x, y) : x^{2} + y^{2} = 1\}$ A={(x,y):x^(2)+y^(2)=1}mathrm{A}=left{(mathrm{x}, mathrm{y}): mathrm{x}^2+mathrm{y}^2=1right} $A = {(x, y) : x^{2} + y^{2} = 1}$ under the discrete metric.
8. a) Consider $Z$ $Z$ Zmathbf{Z} $Z$ and let $F_{1}$ $F_{1}$ F_(1)mathcal{F}_1 $F_{1}$ denote the class of subsets of $Z$ $Z$ Zmathbf{Z} $Z$ , given by $F_{1} = {A \subset Z$ $F_{1} = {A \subset Z$ F_(1)={AsubZmathcal{F}_1={mathrm{A} subset mathbf{Z} $F_{1} = {A \subset Z$ : either $A$ $A$ Amathrm{A} $A$ is finite or $A^{c}$ $A^{c}$ A^(c)mathrm{A}^{mathrm{c}} $A^{c}$ is finite }. Check whether $F_{1}$ $F_{1}$ F_(1)mathcal{F}_1 $F_{1}$ is a $σ$ $σ$ sigmasigma $σ$ algebra or not.
b) Let A be any set in $R$ $R$ Rmathbf{R} $R$ , show that $m^{*} (A) = m^{*} (A + x)$ $m^{*} (A) = m^{*} (A + x)$ m^(**)(A)=m^(**)(A+x)m^*(A)=m^*(A+x) $m^{*} (A) = m^{*} (A + x)$ where $m^{*}$ $m^{*}$ m^(**)m^* $m^{*}$ denotes the outer measure.
c) Find the measure of the following sets.
i) $E = ⋂_{n = 1}^{\infty} (a - \frac{1}{n}, b)$ $E = ⋂_{n = 1}^{\infty} (a - \frac{1}{n}, b)$ quad E=nnn_(n=1)^(oo)(a-(1)/(n),b)quad E=bigcap_{n=1}^{infty}left(a-frac{1}{n}, bright) $E = ⋂_{n = 1}^{\infty} (a - \frac{1}{n}, b)$
ii) $E = Q \cup {1, 2, 3, 4}$ $E = Q \cup {1, 2, 3, 4}$ E=Quu{1,2,3,4}mathrm{E}=mathbf{Q} cup{1,2,3,4} $E = Q \cup {1, 2, 3, 4}$
iii) $E =] 5, 7 [\cup [7, 7.5]$ $E =] 5, 7 [\cup [7, 7.5]$ E=]5,7[uu[7,7.5]mathrm{E}=] 5,7[cup[7,7.5] $E =] 5, 7 [\cup [7, 7.5]$ .
9. a) Show that if $f$ $f$ ff $f$ is measurable, then the function $f^{a} (x)$ $f^{a} (x)$ f^(a)(x)f^a(x) $f^{a} (x)$ given by
$f^{a} (x) = {\begin{array}{cc} a & if f (x) > a \\ f (x) & if f (x) \leq a \end{array}$ $f^{a} (x) = \{\begin{array}{cc} a & if f (x) > a \\ f (x) & if f (x) \leq a \end{array}$ f^(a)(x)={[a,” if “f(x) > a],[f(x),” if “f(x) <= a]:}f^a(x)=left{begin{array}{cc}
a & text { if } f(x)>a \
f(x) & text { if } f(x) leq a
end{array}right. $f^{a} (x) = {\begin{array}{cc} a & if f (x) > a \\ f (x) & if f (x) \leq a \end{array}$
is also measurable.
b) Verify Bounded Convergence Theorem for the sequence of functions ${f_{n}}$ $\{f_{n}\}$ {f_(n)}left{f_nright} ${f_{n}}$ where
$f_{n} (x) = \frac{1}{(1 + x / n)^{n}}, 0 \leq x \leq 1, n \in N$ $f_{n} (x) = \frac{1}{(1 + x / n)^{n}}, 0 \leq x \leq 1, n \in N$ f_(n)(x)=(1)/((1+x//n)^(n)),0 <= x <= 1,ninNmathrm{f}_{mathrm{n}}(mathrm{x})=frac{1}{(1+mathrm{x} / mathrm{n})^{mathrm{n}}}, 0 leq mathrm{x} leq 1, mathrm{n} in mathbf{N} $f_{n} (x) = \frac{1}{(1 + x / n)^{n}}, 0 \leq x \leq 1, n \in N$
c) Find the fourier series of the function $f$ $f$ ff $f$ defined by
$f (x) = {\begin{matrix} - x^{2}, - π < x \leq 0 \\ x^{2}, 0 < x < π \end{matrix}$ $f (x) = \{\begin{matrix} - x^{2}, - π < x \leq 0 \\ x^{2}, 0 < x < π \end{matrix}$ f(x)={[-x^(2)”,”-pi < x <= 0],[x^(2)”,”0 < x < pi]:}f(x)=left{begin{array}{c}
-x^2,-pi<x leq 0 \
x^2, 0<x<pi
end{array}right. $f (x) = {\begin{matrix} - x^{2}, - π < x \leq 0 \\ x^{2}, 0 < x < π \end{matrix}$

State whether the following statements are True or False. Justify your answers.
a) The sequence ${(\frac{1}{n}, \frac{1}{n}) : n \in N}$ $\{(\frac{1}{n}, \frac{1}{n}) : n \in N\}$ {((1)/(n),(1)/(n)):ninN}left{left(frac{1}{mathrm{n}}, frac{1}{mathrm{n}}right): mathrm{n} in mathbf{N}right} ${(\frac{1}{n}, \frac{1}{n}) : n \in N}$ is convergent in $R^{2}$ $R^{2}$ R^(2)mathbf{R}^2 $R^{2}$ under the discrete metric on $R^{2}$ $R^{2}$ R^(2)mathbf{R}^2 $R^{2}$ .
b) A subset in a metric space is compact if it is closed.
c) Continuous image of a path connected space is path connected.
d) The second derivative of a linear map from $R^{n}$ $R^{n}$ R^(n)mathbf{R}^n $R^{n}$ to $R^{m}$ $R^{m}$ R^(m)mathbf{R}^m $R^{m}$ never vanishes.
e) If $\int_{A} f d m = \int_{A} g d m$ $\int_{A} f d m = \int_{A} g d m$ int_(A)fdm=int_(A)gdmint_{mathrm{A}} mathrm{fdm}=int_{mathrm{A}} mathrm{gdm} $\int_{A} f d m = \int_{A} g d m$ for all $A \in M$ $A \in M$ Ain Mmathrm{A} in boldsymbol{M} $A \in M$ , then $f = g$ $f = g$ f=gmathrm{f}=mathrm{g} $f = g$ .

[stray-random categories=trigonometry]

MMT-004 Sample Solution 2024

mmt-004-solved-assignment-2024-ss-020cab3d-1c01-486f-9bdf-7506d86b97ee

MMT-004 Solved Assignment 2024 SS

a) Let $X = C [0, 1]$ $X = C [0, 1]$ X=C[0,1]mathrm{X}=mathrm{C}[0,1] $X = C [0, 1]$ . Define $d : X \times X \to R$ $d : X \times X \to R$ d:X xx X rarrRd: X times X rightarrow mathbf{R} $d : X \times X \to R$ by $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ d(f,g)=int_(0)^(1)|f(t)-g(t)|dt,f,ginXmathrm{d}(mathrm{f}, mathrm{g})=int_0^1|mathrm{f}(mathrm{t})-mathrm{g}(mathrm{t})| mathrm{dt}, mathrm{f}, mathrm{g} in mathrm{X} $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t, f, g \in X$ where the integral is the Riemann integral. Show that $d$ $d$ dd $d$ is a metric on $X$ $X$ XX $X$ . Find $d (f, g)$ $d (f, g)$ d(f,g)d(f, g) $d (f, g)$ where $f (x) = 4 x$ $f (x) = 4 x$ f(x)=4xf(x)=4 x $f (x) = 4 x$ and $g (x) = x^{3}, x \in [0, 1]$ $g (x) = x^{3}, x \in [0, 1]$ g(x)=x^(3),x in[0,1]g(x)=x^3, x in[0,1] $g (x) = x^{3}, x \in [0, 1]$ .

Answer:
To show that $d$ $d$ dd $d$ is a metric on $X$ $X$ XX $X$ , we need to verify the following properties for all $f, g, h \in X$ $f, g, h \in X$ f,g,h in Xf, g, h in X $f, g, h \in X$ :

Non-negativity: $d (f, g) \geq 0$ $d (f, g) \geq 0$ d(f,g) >= 0d(f, g) geq 0 $d (f, g) \geq 0$
Identity of indiscernibles: $d (f, g) = 0$ $d (f, g) = 0$ d(f,g)=0d(f, g) = 0 $d (f, g) = 0$ if and only if $f = g$ $f = g$ f=gf = g $f = g$
Symmetry: $d (f, g) = d (g, f)$ $d (f, g) = d (g, f)$ d(f,g)=d(g,f)d(f, g) = d(g, f) $d (f, g) = d (g, f)$
Triangle inequality: $d (f, h) \leq d (f, g) + d (g, h)$ $d (f, h) \leq d (f, g) + d (g, h)$ d(f,h) <= d(f,g)+d(g,h)d(f, h) leq d(f, g) + d(g, h) $d (f, h) \leq d (f, g) + d (g, h)$

Let’s verify each property:

Non-negativity:
For any $f, g \in X$ $f, g \in X$ f,g in Xf, g in X $f, g \in X$ , the absolute value function $| \cdot |$ $| \cdot |$ |*||cdot| $| \cdot |$ ensures that $| f (t) - g (t) | \geq 0$ $| f (t) - g (t) | \geq 0$ |f(t)-g(t)| >= 0|f(t) – g(t)| geq 0 $| f (t) - g (t) | \geq 0$ for all $t \in [0, 1]$ $t \in [0, 1]$ t in[0,1]t in [0, 1] $t \in [0, 1]$ . Therefore, the integral $\int_{0}^{1} | f (t) - g (t) | d t$ $\int_{0}^{1} | f (t) - g (t) | d t$ int_(0)^(1)|f(t)-g(t)|dtint_0^1 |f(t) – g(t)| , dt $\int_{0}^{1} | f (t) - g (t) | d t$ is also non-negative. Hence, $d (f, g) \geq 0$ $d (f, g) \geq 0$ d(f,g) >= 0d(f, g) geq 0 $d (f, g) \geq 0$ .
Identity of indiscernibles:
If $f = g$ $f = g$ f=gf = g $f = g$ , then $f (t) - g (t) = 0$ $f (t) - g (t) = 0$ f(t)-g(t)=0f(t) – g(t) = 0 $f (t) - g (t) = 0$ for all $t \in [0, 1]$ $t \in [0, 1]$ t in[0,1]t in [0, 1] $t \in [0, 1]$ , so $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} 0 d t = 0$ $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} 0 d t = 0$ d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)0dt=0d(f, g) = int_0^1 |f(t) – g(t)| , dt = int_0^1 0 , dt = 0 $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} 0 d t = 0$ .
Conversely, if $d (f, g) = 0$ $d (f, g) = 0$ d(f,g)=0d(f, g) = 0 $d (f, g) = 0$ , then $\int_{0}^{1} | f (t) - g (t) | d t = 0$ $\int_{0}^{1} | f (t) - g (t) | d t = 0$ int_(0)^(1)|f(t)-g(t)|dt=0int_0^1 |f(t) – g(t)| , dt = 0 $\int_{0}^{1} | f (t) - g (t) | d t = 0$ . Since the integrand is non-negative, it must be zero almost everywhere, implying that $f (t) = g (t)$ $f (t) = g (t)$ f(t)=g(t)f(t) = g(t) $f (t) = g (t)$ for almost all $t \in [0, 1]$ $t \in [0, 1]$ t in[0,1]t in [0, 1] $t \in [0, 1]$ . Since $f$ $f$ ff $f$ and $g$ $g$ gg $g$ are continuous, they must be equal everywhere on $[0, 1]$ $[0, 1]$ [0,1][0, 1] $[0, 1]$ , so $f = g$ $f = g$ f=gf = g $f = g$ .
Symmetry:
By the properties of the absolute value function, $| f (t) - g (t) | = | g (t) - f (t) |$ $| f (t) - g (t) | = | g (t) - f (t) |$ |f(t)-g(t)|=|g(t)-f(t)||f(t) – g(t)| = |g(t) – f(t)| $| f (t) - g (t) | = | g (t) - f (t) |$ for all $t \in [0, 1]$ $t \in [0, 1]$ t in[0,1]t in [0, 1] $t \in [0, 1]$ . Therefore, $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} | g (t) - f (t) | d t = d (g, f)$ $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} | g (t) - f (t) | d t = d (g, f)$ d(f,g)=int_(0)^(1)|f(t)-g(t)|dt=int_(0)^(1)|g(t)-f(t)|dt=d(g,f)d(f, g) = int_0^1 |f(t) – g(t)| , dt = int_0^1 |g(t) – f(t)| , dt = d(g, f) $d (f, g) = \int_{0}^{1} | f (t) - g (t) | d t = \int_{0}^{1} | g (t) - f (t) | d t = d (g, f)$ .
Triangle inequality:
For any $f, g, h \in X$ $f, g, h \in X$ f,g,h in Xf, g, h in X $f, g, h \in X$ and for all $t \in [0, 1]$ $t \in [0, 1]$ t in[0,1]t in [0, 1] $t \in [0, 1]$ , we have $| f (t) - h (t) | = | f (t) - g (t) + g (t) - h (t) | \leq | f (t) - g (t) | + | g (t) - h (t) |$ $| f (t) - h (t) | = | f (t) - g (t) + g (t) - h (t) | \leq | f (t) - g (t) | + | g (t) - h (t) |$ |f(t)-h(t)|=|f(t)-g(t)+g(t)-h(t)| <= |f(t)-g(t)|+|g(t)-h(t)||f(t) – h(t)| = |f(t) – g(t) + g(t) – h(t)| leq |f(t) – g(t)| + |g(t) – h(t)| $| f (t) - h (t) | = | f (t) - g (t) + g (t) - h (t) | \leq | f (t) - g (t) | + | g (t) - h (t) |$ by the triangle inequality for real numbers. Integrating both sides over $[0, 1]$ $[0, 1]$ [0,1][0, 1] $[0, 1]$ gives:
$\int_{0}^{1} | f (t) - h (t) | d t \leq \int_{0}^{1} (| f (t) - g (t) | + | g (t) - h (t) |) d t = \int_{0}^{1} | f (t) - g (t) | d t + \int_{0}^{1} | g (t) - h (t) | d t = d (f, g) + d (g, h)$ $\int_{0}^{1} | f (t) - h (t) | d t \leq \int_{0}^{1} (| f (t) - g (t) | + | g (t) - h (t) |) d t = \int_{0}^{1} | f (t) - g (t) | d t + \int_{0}^{1} | g (t) - h (t) | d t = d (f, g) + d (g, h)$ int_(0)^(1)|f(t)-h(t)|dt <= int_(0)^(1)(|f(t)-g(t)|+|g(t)-h(t)|)dt=int_(0)^(1)|f(t)-g(t)|dt+int_(0)^(1)|g(t)-h(t)|dt=d(f,g)+d(g,h)int_0^1 |f(t) – h(t)| , dt leq int_0^1 (|f(t) – g(t)| + |g(t) – h(t)|) , dt = int_0^1 |f(t) – g(t)| , dt + int_0^1 |g(t) – h(t)| , dt = d(f, g) + d(g, h) $\int_{0}^{1} | f (t) - h (t) | d t \leq \int_{0}^{1} (| f (t) - g (t) | + | g (t) - h (t) |) d t = \int_{0}^{1} | f (t) - g (t) | d t + \int_{0}^{1} | g (t) - h (t) | d t = d (f, g) + d (g, h)$
Thus, $d (f, h) \leq d (f, g) + d (g, h)$ $d (f, h) \leq d (f, g) + d (g, h)$ d(f,h) <= d(f,g)+d(g,h)d(f, h) leq d(f, g) + d(g, h) $d (f, h) \leq d (f, g) + d (g, h)$ .

Since all four properties are satisfied, $d$ $d$ dd $d$ is a metric on $X$ $X$ XX $X$ .
Next, let’s find $d (f, g)$ $d (f, g)$ d(f,g)d(f, g) $d (f, g)$ where $f (x) = 4 x$ $f (x) = 4 x$ f(x)=4xf(x) = 4x $f (x) = 4 x$ and $g (x) = x^{3}$ $g (x) = x^{3}$ g(x)=x^(3)g(x) = x^3 $g (x) = x^{3}$ for $x \in [0, 1]$ $x \in [0, 1]$ x in[0,1]x in [0, 1] $x \in [0, 1]$ :
$d (f, g) = \int_{0}^{1} | 4 x - x^{3} | d x$ $d (f, g) = \int_{0}^{1} | 4 x - x^{3} | d x$ d(f,g)=int_(0)^(1)|4x-x^(3)|dxd(f, g) = int_0^1 |4x – x^3| , dx $d (f, g) = \int_{0}^{1} | 4 x - x^{3} | d x$
To evaluate this integral, we can split it into regions where the integrand is positive or negative:

For $0 \leq x \leq 2$ $0 \leq x \leq 2$ 0 <= x <= 20 leq x leq 2 $0 \leq x \leq 2$ , $4 x \geq x^{3}$ $4 x \geq x^{3}$ 4x >= x^(3)4x geq x^3 $4 x \geq x^{3}$ , so $| 4 x - x^{3} | = 4 x - x^{3}$ $| 4 x - x^{3} | = 4 x - x^{3}$ |4x-x^(3)|=4x-x^(3)|4x – x^3| = 4x – x^3 $| 4 x - x^{3} | = 4 x - x^{3}$ .
For $2 < x \leq 1$ $2 < x \leq 1$ 2 < x <= 12 < x leq 1 $2 < x \leq 1$ , $4 x < x^{3}$ $4 x < x^{3}$ 4x < x^(3)4x < x^3 $4 x < x^{3}$ , so $| 4 x - x^{3} | = x^{3} - 4 x$ $| 4 x - x^{3} | = x^{3} - 4 x$ |4x-x^(3)|=x^(3)-4x|4x – x^3| = x^3 – 4x $| 4 x - x^{3} | = x^{3} - 4 x$ .

Thus, we can rewrite the integral as:
$d (f, g) = \int_{0}^{2} (4 x - x^{3}) d x + \int_{2}^{1} (x^{3} - 4 x) d x$ $d (f, g) = \int_{0}^{2} (4 x - x^{3}) d x + \int_{2}^{1} (x^{3} - 4 x) d x$ d(f,g)=int_(0)^(2)(4x-x^(3))dx+int_(2)^(1)(x^(3)-4x)dxd(f, g) = int_0^{2} (4x – x^3) , dx + int_{2}^1 (x^3 – 4x), dx $d (f, g) = \int_{0}^{2} (4 x - x^{3}) d x + \int_{2}^{1} (x^{3} - 4 x) d x$
Now, we can calculate these integrals:
$\begin{aligned} d (f, g) & = {[2 x^{2} - \frac{1}{4} x^{4}]}_{0}^{2} + {[\frac{1}{4} x^{4} - 2 x^{2}]}_{2}^{1} \\ = (2 (2)^{2} - \frac{1}{4} (2)^{4}) - (0 - 0) + (\frac{1}{4} (1)^{4} - 2 (1)^{2}) - (\frac{1}{4} (2)^{4} - 2 (2)^{2}) \\ = (8 - 4) + (\frac{1}{4} - 2) - (4 - 8) \\ = 4 - \frac{7}{4} - (- 4) \\ = \frac{16}{4} - \frac{7}{4} + \frac{16}{4} \\ = \frac{25}{4} \\ = 6.25 \end{aligned}$ $\begin{aligned} d (f, g) = {[2 x^{2} - \frac{1}{4} x^{4}]}_{0}^{2} + {[\frac{1}{4} x^{4} - 2 x^{2}]}_{2}^{1} \\ = (2 (2)^{2} - \frac{1}{4} (2)^{4}) - (0 - 0) + (\frac{1}{4} (1)^{4} - 2 (1)^{2}) - (\frac{1}{4} (2)^{4} - 2 (2)^{2}) \\ = (8 - 4) + (\frac{1}{4} - 2) - (4 - 8) \\ = 4 - \frac{7}{4} - (- 4) \\ = \frac{16}{4} - \frac{7}{4} + \frac{16}{4} \\ = \frac{25}{4} \\ = 6.25 \end{aligned}$ {:[d(f”,”g)=[2x^(2)-(1)/(4)x^(4)]_(0)^(2)+[(1)/(4)x^(4)-2x^(2)]_(2)^(1)],[=(2(2)^(2)-(1)/(4)(2)^(4))-(0-0)+((1)/(4)(1)^(4)-2(1)^(2))-((1)/(4)(2)^(4)-2(2)^(2))],[=(8-4)+((1)/(4)-2)-(4-8)],[=4-(7)/(4)-(-4)],[=(16)/(4)-(7)/(4)+(16)/(4)],[=(25)/(4)],[=6.25]:}begin{align*}
d(f, g) &= left[2x^2 – frac{1}{4}x^4right]_0^{2} + left[frac{1}{4}x^4 – 2x^2right]_{2}^1 \
&= left(2(2)^2 – frac{1}{4}(2)^4right) – left(0 – 0right) + left(frac{1}{4}(1)^4 – 2(1)^2right) – left(frac{1}{4}(2)^4 – 2(2)^2right) \
&= left(8 – 4right) + left(frac{1}{4} – 2right) – left(4 – 8right) \
&= 4 – frac{7}{4} – (-4) \
&= frac{16}{4} – frac{7}{4} + frac{16}{4} \
&= frac{25}{4} \
&= 6.25
end{align*} $\begin{aligned} d (f, g) & = {[2 x^{2} - \frac{1}{4} x^{4}]}_{0}^{2} + {[\frac{1}{4} x^{4} - 2 x^{2}]}_{2}^{1} \\ = (2 (2)^{2} - \frac{1}{4} (2)^{4}) - (0 - 0) + (\frac{1}{4} (1)^{4} - 2 (1)^{2}) - (\frac{1}{4} (2)^{4} - 2 (2)^{2}) \\ = (8 - 4) + (\frac{1}{4} - 2) - (4 - 8) \\ = 4 - \frac{7}{4} - (- 4) \\ = \frac{16}{4} - \frac{7}{4} + \frac{16}{4} \\ = \frac{25}{4} \\ = 6.25 \end{aligned}$
Therefore, $d (f, g) = 6.25$ $d (f, g) = 6.25$ d(f,g)=6.25d(f, g) = 6.25 $d (f, g) = 6.25$ .
b) Let ( $X, d)$ $X, d)$ X,d)X, d) $X, d)$ be a metric space and $a \in X$ $a \in X$ a in Xa in X $a \in X$ be a fixed point of $X$ $X$ XX $X$ . Show that the function $f_{a} : X \to R$ $f_{a} : X \to R$ f_(a):X rarrRf_a: X rightarrow mathbf{R} $f_{a} : X \to R$ given by $f_{a} (x) = d (x, a)$ $f_{a} (x) = d (x, a)$ f_(a)(x)=d(x,a)mathrm{f}_{mathrm{a}}(mathrm{x})=mathrm{d}(mathrm{x}, mathrm{a}) $f_{a} (x) = d (x, a)$ is continuous. Is it uniformly continuous? Justify you answer.
Answer:
To show that the function $f_{a} : X \to R$ $f_{a} : X \to R$ f_(a):X rarrRf_a: X rightarrow mathbf{R} $f_{a} : X \to R$ given by $f_{a} (x) = d (x, a)$ $f_{a} (x) = d (x, a)$ f_(a)(x)=d(x,a)f_a(x) = d(x, a) $f_{a} (x) = d (x, a)$ is continuous, we need to show that for every $ϵ > 0$ $ϵ > 0$ epsilon > 0epsilon > 0 $ϵ > 0$ , there exists a $δ > 0$ $δ > 0$ delta > 0delta > 0 $δ > 0$ such that for all $x, y \in X$ $x, y \in X$ x,y in Xx, y in X $x, y \in X$ , if $d (x, y) < δ$ $d (x, y) < δ$ d(x,y) < deltad(x, y) < delta $d (x, y) < δ$ , then $| f_{a} (x) - f_{a} (y) | < ϵ$ $| f_{a} (x) - f_{a} (y) | < ϵ$ |f_(a)(x)-f_(a)(y)| < epsilon|f_a(x) – f_a(y)| < epsilon $| f_{a} (x) - f_{a} (y) | < ϵ$ .
Let $ϵ > 0$ $ϵ > 0$ epsilon > 0epsilon > 0 $ϵ > 0$ be given. Set $δ = ϵ$ $δ = ϵ$ delta=epsilondelta = epsilon $δ = ϵ$ . Then, for any $x, y \in X$ $x, y \in X$ x,y in Xx, y in X $x, y \in X$ such that $d (x, y) < δ$ $d (x, y) < δ$ d(x,y) < deltad(x, y) < delta $d (x, y) < δ$ , we have:
$\begin{aligned} | f_{a} (x) - f_{a} (y) | & = | d (x, a) - d (y, a) | \\ \leq d (x, y) (by the triangle inequality) \\ < δ \\ = ϵ \end{aligned}$ $\begin{aligned} | f_{a} (x) - f_{a} (y) | = | d (x, a) - d (y, a) | \\ \leq d (x, y) (by the triangle inequality) \\ < δ \\ = ϵ \end{aligned}$ {:[|f_(a)(x)-f_(a)(y)|=|d(x”,”a)-d(y”,”a)|],[ <= d(x”,”y)quad(by the triangle inequality)],[ < delta],[=epsilon]:}begin{align*}
|f_a(x) – f_a(y)| &= |d(x, a) – d(y, a)| \
&leq d(x, y) quad text{(by the triangle inequality)} \
&< delta \
&= epsilon
end{align*} $\begin{aligned} | f_{a} (x) - f_{a} (y) | & = | d (x, a) - d (y, a) | \\ \leq d (x, y) (by the triangle inequality) \\ < δ \\ = ϵ \end{aligned}$
Hence, $f_{a}$ $f_{a}$ f_(a)f_a $f_{a}$ is continuous.
Now, let’s consider whether $f_{a}$ $f_{a}$ f_(a)f_a $f_{a}$ is uniformly continuous. A function is uniformly continuous if for every $ϵ > 0$ $ϵ > 0$ epsilon > 0epsilon > 0 $ϵ > 0$ , there exists a $δ > 0$ $δ > 0$ delta > 0delta > 0 $δ > 0$ such that for all $x, y \in X$ $x, y \in X$ x,y in Xx, y in X $x, y \in X$ , if $d (x, y) < δ$ $d (x, y) < δ$ d(x,y) < deltad(x, y) < delta $d (x, y) < δ$ , then $| f_{a} (x) - f_{a} (y) | < ϵ$ $| f_{a} (x) - f_{a} (y) | < ϵ$ |f_(a)(x)-f_(a)(y)| < epsilon|f_a(x) – f_a(y)| < epsilon $| f_{a} (x) - f_{a} (y) | < ϵ$ , where $δ$ $δ$ deltadelta $δ$ does not depend on the choice of $x$ $x$ xx $x$ or $y$ $y$ yy $y$ .
In metric spaces, continuity and uniform continuity are not always equivalent. However, if the metric space $X$ $X$ XX $X$ is compact, then every continuous function from $X$ $X$ XX $X$ to $R$ $R$ Rmathbf{R} $R$ is uniformly continuous. This is a consequence of the Heine-Cantor theorem.
If the metric space $X$ $X$ XX $X$ is not compact, we cannot guarantee that $f_{a}$ $f_{a}$ f_(a)f_a $f_{a}$ is uniformly continuous. For example, consider the metric space $(R, d)$ $(R, d)$ (R,d)(mathbf{R}, d) $(R, d)$ where $d (x, y) = | x - y |$ $d (x, y) = | x - y |$ d(x,y)=|x-y|d(x, y) = |x – y| $d (x, y) = | x - y |$ is the standard metric on $R$ $R$ Rmathbf{R} $R$ . The function $f_{a} (x) = | x - a |$ $f_{a} (x) = | x - a |$ f_(a)(x)=|x-a|f_a(x) = |x – a| $f_{a} (x) = | x - a |$ is continuous but not uniformly continuous on $R$ $R$ Rmathbf{R} $R$ .
In conclusion, the function $f_{a}$ $f_{a}$ f_(a)f_a $f_{a}$ is continuous, but it is not necessarily uniformly continuous unless the metric space $X$ $X$ XX $X$ has additional properties, such as compactness.

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