State whether the following statements are true or false. Justify your answers.
a) The outer measure m^(**)m^* of the set A={x inR:x^(2)=1}uu[-3,2]A=\left\{x \in \mathbb{R}: x^2=1\right\} \cup[-3,2] is 0 .
b) A finite subset of a metric space is totally bounded.
c) A connected subspace in a metric space which in not properly contained in any other connected subspace is always open.
d) The surface given by the equation x+y+z-sin(xyz)=0x+y+z-\sin (x y z)=0 can also be described by an equation of the form z=f(x,y)z=f(x, y) in a neighbourhood of the point (0,0)(0,0).
e) A real valued function ff on [a,b][a, b] is continuous if it is integrable on [a,b][a, b].
a) Find the interior, closure, the set of limit points and the boundary of the set
a) Let f:R^(3)\\{(0,0,0)}rarrR^(3)\\{(0,0,0)}f: \mathbb{R}^3 \backslash\{(0,0,0)\} \rightarrow \mathbb{R}^3 \backslash\{(0,0,0)\} be given by
f(x,y,z)=((x)/(x^(2)+y^(2)+3^(2)),(y)/(x^(2)+y^(2)+z^(2)),(z)/(x^(2)+y^(2)+z^(2)))f(x, y, z)=\left(\frac{x}{x^2+y^2+3^2}, \frac{y}{x^2+y^2+z^2}, \frac{z}{x^2+y^2+z^2}\right)
Show that ff is locally invertible at all points in R^(3)\\{(0,0,0)}\mathbb{R}^3 \backslash\{(0,0,0)\}.
b) For the equation x^(2)+y^(2)+z^(3)=0x^2+y^2+z^3=0, at which points on its solution set, can we assured that there is a neighbourhood of the point in which the surface given by the equation can be described by an equation of the form z=f(x,y)z=f(x, y).
c) Find the Fourier series of f(t)=t^(2)f(t)=t^2 on [-pi,pi][-\pi, \pi].
5. a) Prove that if an open set UU can be written as the union of pariwise disjoint family VV of open connected subsets, then these subsets must be the components of UU. Use this theorem to find the components of the set D uu ED \cup E where
b) Which of the following subsets of R\mathbb{R} are compact w.r.t. the metric given against them. Justify your answer.
i) quad A=(0,1)\quad A=(0,1) in R\mathbb{R} of -R-\mathbb{R} with standard metric.
ii) quad A=[3,4]-R\quad A=[3,4]-\mathbb{R} with discrete metric.
iii) quad{(x,y)inR^(2)quad y > 0}-R^(2)\quad\left\{(x, y) \in \mathbb{R}^2 \quad y>0\right\}-\mathbb{R}^2 with standard metric.
a) If EE is a subset of R\mathbb{R} with standard metric, then show that ( bar(E))^(c)=(E^(C))^(0)(\bar{E})^c=\left(E^C\right)^0.
b) Show that a set AA in a metric space is closed if and only if every convergent sequence in AA converges to a point of AA.
c) Find the interior and closure of the set Q\mathbb{Q} of rationals in R\mathbb{R} with standard metric.
a) Let FF be the function from R^(2)\mathbb{R}^2 to R^(2)\mathbb{R}^2 defined by
F(x,y)=(x^(2)+y^(2),xy)F(x, y)=\left(x^2+y^2, x y\right)
Show that FF is differentiable at (1,2)(1,2). Find the differential matrix of FF.
b) Show that the function ff defined by
f(x,y)={[(xy)/(x^(2)+y^(2))”,”,” if “(x”,”y)!=(0″,”0)],[0,”, if “(x”,”y)=(0″,”0)]:}f(x, y)=\left\{\begin{array}{cc}
\frac{x y}{x^2+y^2}, & \text { if }(x, y) \neq(0,0) \\
0 & \text {, if }(x, y)=(0,0)
\end{array}\right.
is not differentiable at (0,0)(0,0). Does the partial derivatives of ff exists at (0,0)(0,0) ? or any at any other point in R^(2)\mathbb{R}^2 ? Justify your answer.
c) Is the continuous image of a Cauchy sequence a Cauchy sequence? Justify.
8. a) Find the directional derivative of the function f:R^(4)rarrR^(4)f: \mathbb{R}^4 \rightarrow \mathbb{R}^4 defined by
f(x,y,z,w)=(x^(2)y,xyz,x^(2)+y^(2),zw^(2))f(x, y, z, w)=\left(x^2 y, x y z, x^2+y^2, z w^2\right)
at the point (1,2,-1,-2)(1,2,-1,-2) in the direction v=(1,0,-2,2)v=(1,0,-2,2).
b) Suppose that f:RrarrR^(2)f: \mathbb{R} \rightarrow \mathbb{R}^2 is given by f(t)=(t,t^(2))f(t)=\left(t, t^2\right) and g:R^(2)rarrR^(3)g: \mathbb{R}^2 \rightarrow \mathbb{R}^3 is given by g(x,y)=(x^(2),xy,y^(2)-x^(2))g(x, y)=\left(x^2, x y, y^2-x^2\right). Compute the derivative of g@fg \circ f.
c) Find B[2,(1)/(2)]B\left[2, \frac{1}{2}\right] in R\mathbb{R} where dd is the metric given by d(x,y)=min{1,|x-y|}d(x, y)=\min \{1,|x-y|\}.
9. a) Give an example of a family f_(1)f_1 of subsets of a set XX which has finite intersection property. Justify your choice of example.
b) Verify the hypothesis and conclusions of the Fatou’s lemma for the sequence {f_(n)}\left\{f_n\right\} given by
{:[f_(n)(x)=2n” for “x in((1)/(2n),(1)/(n))],[=0” for “x in(0,(1)/(2n))uu((1)/(n),1)]:}\begin{aligned}
f_n(x) & =2 n \text { for } x \in\left(\frac{1}{2 n}, \frac{1}{n}\right) \\
& =0 \text { for } x \in\left(0, \frac{1}{2 n}\right) \cup\left(\frac{1}{n}, 1\right)
\end{aligned}
a) Let (X,d)(X, d) be a metric space and AA be a subset of XX. Show that bdy(A)=Qb d y(A)=Q if and only if AA is both open and closed.
b) Give an example of an algebra which is not a sigma\sigma – algebra. Justify your choice of examples.
c) If EE is a measurable set and ff is a simple function such that a <= f(x) <= b AA x in Ea \leq f(x) \leq b \forall x \in E, show that
am(E) <= int _(E)fdm <= m(E)ba m(E) \leq \int_E f d m \leq m(E) b
Answer:
Question:-1
State whether the following statements are true or false. Justify your answers.
a) The outer measure m^(**)m^* of the set A={x inR:x^(2)=1}uu[-3,2]A=\left\{x \in \mathbb{R}: x^2=1\right\} \cup[-3,2] is 0.
b) A finite subset of a metric space is totally bounded.
c) A connected subspace in a metric space which is not properly contained in any other connected subspace is always open.
d) The surface given by the equation x+y+z-sin(xyz)=0x+y+z-\sin (x y z)=0 can also be described by an equation of the form z=f(x,y)z=f(x, y) in a neighbourhood of the point (0,0)(0,0).
e) A real valued function ff on [a,b][a, b] is continuous if it is integrable on [a,b][a, b].
Answer:
a) The outer measure m^(**)m^* of the set A={x inR:x^(2)=1}uu[-3,2]A=\left\{x \in \mathbb{R}: x^2=1\right\} \cup[-3,2] is 0.
Let’s evaluate the statement: "The outer measure m^(**)m^* of the set A={x inR:x^(2)=1}uu[-3,2]A = \{x \in \mathbb{R} : x^2 = 1\} \cup [-3, 2] is 0."
First, define the set AA. The condition x^(2)=1x^2 = 1 implies x=1x = 1 or x=-1x = -1, so {x inR:x^(2)=1}={-1,1}\{x \in \mathbb{R} : x^2 = 1\} = \{-1, 1\}. Then, A={-1,1}uu[-3,2]A = \{-1, 1\} \cup [-3, 2]. Since {-1,1}\{-1, 1\} is a subset of [-3,2][-3, 2] (as -1in[-3,2]-1 \in [-3, 2] and 1in[-3,2]1 \in [-3, 2]), the union simplifies to A=[-3,2]A = [-3, 2].
Next, compute the outer measure m^(**)(A)m^*(A). For a bounded interval [a,b][a, b], the Lebesgue outer measure is the length of the interval: m^(**)([a,b])=b-am^*([a, b]) = b – a. Here, A=[-3,2]A = [-3, 2], so m^(**)([-3,2])=2-(-3)=5m^*([-3, 2]) = 2 – (-3) = 5.
The statement claims m^(**)(A)=0m^*(A) = 0, but we calculated m^(**)(A)=5m^*(A) = 5. To confirm, note that outer measure is non-negative and zero only for sets that can be covered by intervals of arbitrarily small total length (e.g., finite or countable sets). The interval [-3,2][-3, 2] has length 5 and cannot be covered by intervals of total length less than 5, so m^(**)(A)!=0m^*(A) \neq 0.
Thus, the statement is false. Proof: A=[-3,2]A = [-3, 2], and m^(**)([-3,2])=5 > 0m^*([-3, 2]) = 5 > 0.
b) A finite subset of a metric space is totally bounded.
Let’s evaluate the statement: "A finite subset of a metric space is totally bounded."
In a metric space, a set SS is totally bounded if, for every epsilon > 0\epsilon > 0, it can be covered by finitely many balls of radius epsilon\epsilon. Consider a finite subset S={x_(1),x_(2),dots,x_(n)}S = \{x_1, x_2, \ldots, x_n\} of a metric space (X,d)(X, d). For any epsilon > 0\epsilon > 0, we can cover SS with balls centered at each point: take the collection B(x_(1),epsilon),B(x_(2),epsilon),dots,B(x_(n),epsilon)B(x_1, \epsilon), B(x_2, \epsilon), \ldots, B(x_n, \epsilon). Each point x_(i)in B(x_(i),epsilon)x_i \in B(x_i, \epsilon) (since d(x_(i),x_(i))=0 < epsilond(x_i, x_i) = 0 < \epsilon), so S subeuuu_(i=1)^(n)B(x_(i),epsilon)S \subseteq \bigcup_{i=1}^n B(x_i, \epsilon). This is a finite cover of SS by epsilon\epsilon-balls, satisfying the definition of total boundedness.
Could it be false? In a metric space, a finite set has no limit points beyond itself, and its discrete nature ensures a finite cover always works, regardless of the metric. For example, in R\mathbb{R} with S={1,2}S = \{1, 2\}, for epsilon=0.1\epsilon = 0.1, balls B(1,0.1)B(1, 0.1) and B(2,0.1)B(2, 0.1) cover SS.
Thus, the statement is true. Proof: A finite set S={x_(1),dots,x_(n)}S = \{x_1, \ldots, x_n\} can be covered by nn balls of radius epsilon\epsilon, one centered at each point, for any epsilon > 0\epsilon > 0.
c) A connected subspace in a metric space which is not properly contained in any other connected subspace is always open.
Let’s evaluate the statement: "A connected subspace in a metric space which is not properly contained in any other connected subspace is always open."
Consider a metric space (X,d)(X, d) and a connected subspace S sube XS \subseteq X that is maximal, meaning no connected subspace of XX properly contains SS. We need to determine if SS must be open.
Take X=RX = \mathbb{R} with the standard metric, and let S=[0,1]S = [0, 1]. First, check if SS is connected: the interval [0,1][0, 1] is connected in R\mathbb{R} because any disconnection into two nonempty open sets would contradict the intermediate value theorem for continuous functions on [0,1][0, 1]. Next, is SS maximal? Suppose T sup[0,1]T \supset [0, 1] is connected and T!=[0,1]T \neq [0, 1]. If TT includes a point x > 1x > 1, say T=[0,a]T = [0, a] with a > 1a > 1, this is still connected (intervals in R\mathbb{R} are connected). Thus, [0,1][0, 1] is not maximal, as [0,2][0, 2] is a larger connected set containing it.
Now consider S=RS = \mathbb{R} itself. It’s connected (as a complete metric space with no gaps), and maximal since no proper superset of R\mathbb{R} exists in R\mathbb{R}. But R\mathbb{R} is open in R\mathbb{R}, supporting the statement. However, we need a counterexample where SS is maximal and not open. Try X=R^(2)X = \mathbb{R}^2 with the Euclidean metric, and S={(x,0):x in[0,1]}S = \{(x, 0) : x \in [0, 1]\}, the closed unit segment on the x-axis. SS is connected (it’s a line segment). Is it maximal? If T sup ST \supset S is connected and includes a point off the x-axis, say (0,1)(0, 1), then TT could be a path from (0,0)(0, 0) to (0,1)(0, 1) via (1,0)(1, 0), but adding points like (0,1)(0, 1) often allows disconnection (e.g., separate y=0y = 0 and y > 0y > 0). In R^(2)\mathbb{R}^2, S=[0,1]xx{0}S = [0, 1] \times \{0\} is not maximal unless restricted, but consider a discrete metric space X={a,b}X = \{a, b\} with d(a,b)=1d(a, b) = 1, and S={a}S = \{a\}. It’s connected (one point), maximal (adding bb disconnects), and not open (open sets are O/,{a},{b},X\emptyset, \{a\}, \{b\}, X).
A better counterexample: X=[0,1]uu[2,3]X = [0, 1] \cup [2, 3] in R\mathbb{R}, and S=[0,1]S = [0, 1]. SS is connected, and adding any point from [2,3][2, 3] disconnects the subspace (gap between 1 and 2). Thus, S=[0,1]S = [0, 1] is maximal in XX, but not open in XX (e.g., 1 has no open interval around it fully in SS).
The statement is false. Proof: In X=[0,1]uu[2,3]X = [0, 1] \cup [2, 3], S=[0,1]S = [0, 1] is connected, maximal, and not open.
d) The surface given by the equation x+y+z-sin(xyz)=0x+y+z-\sin (x y z)=0 can also be described by an equation of the form z=f(x,y)z=f(x, y) in a neighbourhood of the point (0,0)(0,0).
Let’s evaluate the statement: "The surface given by the equation x+y+z-sin(xyz)=0x + y + z – \sin(xyz) = 0 can also be described by an equation of the form z=f(x,y)z = f(x, y) in a neighborhood of the point (0,0)(0, 0)."
Define the function F(x,y,z)=x+y+z-sin(xyz)F(x, y, z) = x + y + z – \sin(xyz). The surface is F(x,y,z)=0F(x, y, z) = 0. We need to determine if, near (0,0)(0, 0), this implicitly defines zz as a function of xx and yy, i.e., z=f(x,y)z = f(x, y). Check the point: at (0,0,z)(0, 0, z), F(0,0,z)=0+0+z-sin(0*0*z)=z=0F(0, 0, z) = 0 + 0 + z – \sin(0 \cdot 0 \cdot z) = z = 0, so (0,0,0)(0, 0, 0) satisfies the equation.
Use the Implicit Function Theorem: if F=0F = 0 and (del F)/(del z)!=0\frac{\partial F}{\partial z} \neq 0 at (0,0,0)(0, 0, 0), then zz can be solved as a function of xx and yy locally. Compute the partial derivative:
Since (del F)/(del z)=1!=0\frac{\partial F}{\partial z} = 1 \neq 0 at (0,0,0)(0, 0, 0), the theorem applies. Thus, there exists a neighborhood of (0,0)(0, 0) in the xyxy-plane where z=f(x,y)z = f(x, y) satisfies F(x,y,f(x,y))=0F(x, y, f(x, y)) = 0.
The statement is true. Proof: By the Implicit Function Theorem, since (del F)/(del z)=1!=0\frac{\partial F}{\partial z} = 1 \neq 0 at (0,0,0)(0, 0, 0), z=f(x,y)z = f(x, y) exists locally near (0,0)(0, 0).
e) A real valued function ff on [a,b][a, b] is continuous if it is integrable on [a,b][a, b].
Let’s evaluate the statement: "A real-valued function ff on [a,b][a, b] is continuous if it is integrable on [a,b][a, b]."
The statement suggests that integrability (typically Riemann integrability, unless specified otherwise) implies continuity. Consider ff on [0,1][0, 1] defined as f(x)=0f(x) = 0 if xx is irrational and f(x)=1f(x) = 1 if xx is rational. This is a classic example:
Integrability: The set of discontinuities is Qnn[0,1]\mathbb{Q} \cap [0, 1], which has Lebesgue measure zero (rationals are countable). A bounded function on [a,b][a, b] is Riemann integrable if its discontinuities have measure zero, so ff is integrable, with int_(0)^(1)f(x)dx=0\int_0^1 f(x) \, dx = 0 (since irrationals are dense and f=0f = 0 there).
Continuity: At every x in[0,1]x \in [0, 1], ff is discontinuous because rationals and irrationals are dense: for any xx, there are nearby points where f=1f = 1 and others where f=0f = 0, so lim_(y rarr x)f(y)\lim_{y \to x} f(y) doesn’t exist.
Since ff is integrable but not continuous, the implication fails.
The statement is false. Proof: The function f(x)=1f(x) = 1 if xx is rational, 00 if irrational, on [0,1][0, 1] is integrable (discontinuities have measure zero) but discontinuous everywhere.