Chapter 23: Real Analysis and Series of Functions (Set-3)
Which statement correctly describes the interval [a, b)?
A a included only
B Both excluded
C Both included
D b included only
The bracket at a means a is included, while the parenthesis at b means b is excluded. So [a,b) contains all x with a ≤ x < b.
If |x−3| < 2, then x belongs to
A [1, 5]
B (−1, 1)
C (1, 5)
D (5, 7)
|x−3|<2 means the distance from 3 is less than 2, so −2 < x−3 < 2. Adding 3 gives 1 < x < 5, which is (1,5).
Which set has no maximum but has a supremum?
A [0, 1]
B (0, 1)
C {1,2,3}
D (−1, 0]
(0,1) does not contain 1, so no maximum exists. But 1 is still the least upper bound, so the supremum exists and equals 1.
Which set has an infimum not contained in the set?
A [2, 5]
B {0,1,2}
C (2, 5]
D [−1, 3)
In (2,5], the greatest lower bound is 2, but 2 is excluded. So infimum exists and equals 2, yet it is not an element of the set.
If S is nonempty and bounded above in ℝ, then
A sup S exists
B max S exists
C S must be open
D S must be finite
By the completeness (least upper bound) property of ℝ, every nonempty subset of ℝ that is bounded above has a supremum in ℝ, even if it has no maximum.
Which is true about supremum sup S?
A Always in S
B Always smallest element
C Least upper bound
D Greatest lower bound
Supremum is the smallest real number that is an upper bound for S. It may or may not belong to S. It is defined by the order and completeness of ℝ.
Which condition guarantees a set has a maximum?
A Bounded above only
B sup exists only
C Dense in ℝ
D Closed and bounded
In ℝ, closed and bounded sets are compact. A continuous function attains extrema on compact sets; and for sets themselves, compactness helps ensure upper bounds are achieved in many cases.
Archimedean property implies for any x>0, there exists n with
A nx > 1
B nx < 1
C n = x
D n < x always
If x>0, Archimedean property ensures natural numbers grow without bound, so some n makes nx exceed 1. This shows there are no “infinitely small” positive reals in ℝ.
Density of rationals also implies between a<b there exists
A No irrational
B Only integer
C An irrational number
D Only natural number
Between any two real numbers a<b, there exist infinitely many rationals and also infinitely many irrationals. So you can always find an irrational strictly between a and b.
The metric ball centered at a with radius r is
A [a−r, a+r]
B (a, a+r)
C (a−r, a)
D (a−r, a+r)
In the usual metric d(x,a)=|x−a|, the open ball {x:|x−a|<r} equals the open interval (a−r, a+r). It represents points within distance r of a.
If A is bounded above and c>0, then cA is
A Unbounded above
B Bounded above
C Not comparable
D Always empty
If A ≤ M for all a∈A, then ca ≤ cM for c>0. So multiplying by a positive constant preserves boundedness above and scales bounds accordingly.
If A is bounded above and c<0, then cA is
A Bounded above
B Always compact
C Bounded below
D Always open
Multiplying inequalities by a negative number reverses order. If a ≤ M, then ca ≥ cM when c<0. So an upper bound becomes a lower bound after scaling by c<0.
If A is bounded below and B is bounded below, then A∩B is
A Bounded below
B Always unbounded
C Must be empty
D Only finite
If mA is a lower bound of A and mB is a lower bound of B, then min(mA,mB) is a lower bound for A∩B (when intersection is nonempty), so it stays bounded below.
If A ⊆ B and B is bounded above, then A is
A Unbounded above
B Not bounded
C Bounded above
D Dense always
Any upper bound for B also bounds every element of A, since A is a subset. Therefore boundedness passes from a set to its subsets directly.
If sup A exists and A ⊆ B, then
A sup A ≥ sup B
B sup A ≤ sup B
C sup B does not exist
D A must be closed
If A is contained in B, then B can have elements larger than those in A. So the least upper bound of A cannot exceed the least upper bound of B (when both exist).
Which set is bounded but not closed?
A [0,1]
B (−∞,1]
C ℝ
D (0,1)
(0,1) is bounded since it lies between 0 and 1. It is not closed because it does not contain its boundary points 0 and 1, which are limit points.
Which set is closed but not bounded?
A [0,1]
B (0,1)
C [0,∞)
D (−1,2)
[0,∞) contains its limit points, so it is closed in ℝ. But it is not bounded above because it extends to infinity, so it fails boundedness.
If fₙ → f uniformly, then supₓ |fₙ(x)−f(x)|
A Tends to 0
B Equals 1 always
C Tends to infinity
D Oscillates always
Uniform convergence is equivalent to the supremum of the error over the domain going to 0. This captures the idea that the maximum possible error becomes small for large n.
If fₙ → f pointwise, then for each fixed x
A One N fits all
B Error stays constant
C Error goes to 0
D Limit must be continuous
Pointwise convergence means for each specific x, the numeric sequence fₙ(x) converges to f(x). N may depend on x, unlike uniform convergence.
Uniform convergence is stronger because N can be chosen
A Depending on x
B Independent of x
C Only for polynomials
D Only for series
In uniform convergence, given ε>0, there exists one N such that for all n≥N and all x, |fₙ(x)−f(x)|<ε. This global control is the key strength.
If fₙ are bounded and converge uniformly, then f is
A Bounded
B Always unbounded
C Always discontinuous
D Only piecewise
Uniform convergence implies f differs from some bounded fₙ by a small amount uniformly. So f inherits boundedness. Pointwise convergence alone does not guarantee boundedness.
A common sufficient condition for uniform convergence of ∑ fₙ is
A Terms alternate
B Terms increase
C M-test holds
D Only finitely many
If |fₙ(x)| ≤ Mₙ for all x and ∑ Mₙ converges, then ∑ fₙ converges uniformly. This provides an easy domination method without tracking x-dependence.
If ∑ fₙ converges uniformly, then its partial sums S_N are
A Random functions
B Uniformly Cauchy
C Always constant
D Not defined
Uniform convergence means partial sums approach a limit uniformly. Equivalently, the sequence of partial sums is uniformly Cauchy: beyond some N, S_m and S_n are close for all x.
Uniform Cauchy means: for ε>0 there exists N such that
A |S_n(x)|<ε
B S_n(x)=S_m(x)
C Only at one x
D |S_n(x)−S_m(x)|<ε for all x
The uniform Cauchy criterion demands that after some index, all later partial sums are close to each other uniformly in x. This is the key “Cauchy test” for uniform convergence.
Uniform convergence mainly helps to justify
A Swapping limit and continuity
B Swapping limit always
C Making set bounded
D Removing derivatives
If fₙ are continuous and converge uniformly to f, then f is continuous. This is a classic “interchange limit and continuity” result that can fail under only pointwise convergence.
To interchange limit and integral safely, one common requirement is
A Pointwise convergence only
B Series must diverge
C Uniform convergence
D Domain must be empty
Uniform convergence (often along with integrability conditions) allows passing the limit inside the integral. It controls error uniformly, preventing incorrect results that sometimes occur with pointwise convergence.
Why pointwise convergence may fail for integration?
A It changes algebra
B Error not uniform
C Requires monotone
D Needs compactness
With only pointwise convergence, the convergence speed can vary with x. Large errors may remain on parts of the domain, so interchanging limit and integration may produce wrong values.
For a power series, the interval of convergence is
A All x with |x−a|<R plus possible endpoints
B Only one point
C Always (−∞,∞)
D Always empty
A power series converges absolutely for |x−a|<R and diverges for |x−a|>R. At endpoints |x−a|=R, convergence must be checked separately.
If a power series converges at x=b ≠ a, then it converges
A Nowhere else
B Only at endpoints
C For all |x−a|<|b−a|
D Only at x=a
Convergence at one point implies convergence for all points closer to the center a. This follows from the radius concept: if it converges at distance d, then R ≥ d.
If a power series diverges at x=c, then it diverges for
A All closer points
B Only at center
C Only at endpoints
D All farther points
If it diverges at a point with distance d=|c−a|, then R ≤ d. Hence for any x with |x−a|>d, the series must also diverge.
A power series has infinite radius when it converges for
A Only x=0
B Only |x|<1
C All real x
D Only integers
If a power series converges for every real x, its radius of convergence is infinite. Examples include sin x, cos x, and e^x expansions, which work everywhere on ℝ.
Within radius R, the differentiated power series has radius
A Same R
B R−1
C 2R
D 0 only
Differentiating term-by-term inside the radius produces another power series that converges on the same open interval |x−a|<R. The radius remains unchanged.
Within radius R, the integrated power series has radius
A Half of R
B Always 1
C Same R
D Not defined
Term-by-term integration inside the radius yields a new power series with the same radius of convergence. Integration does not reduce the radius, and the resulting series converges on |x−a|<R.
Absolute vs conditional convergence differs because
A Conditional uses |aₙ|
B Absolute means ∑aₙ diverges
C Conditional means finite terms
D Absolute means ∑|aₙ| converges
Absolute convergence requires the series of absolute values to converge. Conditional convergence means ∑aₙ converges but ∑|aₙ| diverges. Absolute convergence gives stronger stability.
A convergent sequence in ℝ has a unique
A Subsequence only
B Limit
C Upper bound only
D Lower bound only
In ℝ, limits are unique: if a sequence converges to L and also to M, then L=M. This follows from the ε-definition and triangle inequality arguments.
If xₙ → L and yₙ → M, then (xₙ + yₙ) →
A LM
B L−M
C L+M
D L/M
Limits respect addition: convergence is preserved under algebraic operations. Since xₙ approaches L and yₙ approaches M, their sum approaches L+M by standard limit laws.
If xₙ → L and c is constant, then cxₙ →
A cL
B L+c
C L/c
D c+1
Multiplying a convergent sequence by a constant scales the limit by the same constant. This follows directly from the ε-definition using |cxₙ−cL|=|c||xₙ−L|.
If xₙ is Cauchy, then for ε>0, there exists N such that
A |xₙ|<ε for n≥N
B xₙ increases after N
C |xₙ−x_m|<ε for m,n≥N
D xₙ equals 0
This is the definition of a Cauchy sequence: terms eventually get arbitrarily close to each other. In ℝ, this condition is enough to guarantee convergence.
A bounded sequence may fail to converge because it can
A Have no terms
B Oscillate
C Be constant
D Be monotone
Boundedness prevents blowing up, but it does not force approaching one value. A classic bounded nonconvergent example is (−1)ⁿ, which oscillates between −1 and 1 forever.
Bolzano–Weierstrass is a key reason bounded sequences relate to
A Polynomial roots
B Matrix inverses
C Differential equations
D Compactness ideas
The theorem that every bounded sequence has a convergent subsequence is closely tied to compactness in ℝ. It explains why closed and bounded sets behave nicely for limits.
Heine–Borel in ℝ states compact sets are exactly
A Dense sets
B Open and bounded
C Closed and bounded
D Closed and unbounded
In ℝ, compactness is equivalent to being closed and bounded. This characterization is very useful for proving existence of maxima/minima and for subsequence arguments.
A set is bounded above if there exists M such that
A x≤M for all x
B x≥M for all x
C x=M for all x
D x≠M for all x
Upper bounded means every element lies at or below some real number M. M is an upper bound. If such an M exists, the set cannot extend to +∞.
Least upper bound means for sup S = u
A u is lower bound
B u is upper bound and any smaller fails
C u must be in S
D u is minimum element
u is an upper bound, and no number less than u can still be an upper bound. This captures “least” upper bound. Supremum can exist even when not achieved.
Greatest lower bound means for inf S = l
A l is upper bound
B l must be in S
C l is lower bound and any larger fails
D l is maximum element
l is a lower bound, and any number greater than l is not a lower bound. That makes l the greatest lower bound. Like supremum, infimum may not belong to S.
A sequence of functions (fₙ) is uniformly bounded if
A |fₙ(x)| ≤ M for all n,x
B |fₙ(x)| ≤ x for all n
C fₙ(x) → 0 only
D Domain is finite
Uniform boundedness means one constant M bounds all functions at all points: |fₙ(x)|≤M for every n and x. This is stronger than each function being bounded separately.
If fₙ → f uniformly and each fₙ is integrable, then typically
A f must be constant
B ∫fₙ diverges
C Integral undefined
D ∫fₙ → ∫f
Uniform convergence (with standard integrability assumptions on a fixed interval) allows interchanging limit and integral: the integrals of fₙ converge to the integral of f because errors are uniformly controlled.
For series ∑ aₙ, the ratio test uses limit of
A aₙ−aₙ₊₁
B |aₙ|+|aₙ₊₁|
C |aₙ₊₁/aₙ|
D aₙ/n
Ratio test looks at L = lim |aₙ₊₁/aₙ| (if it exists). If L<1, series converges absolutely; if L>1, diverges; if L=1, inconclusive.
For series ∑ aₙ, the root test uses
A lim sup √[n]{|aₙ|}
B lim aₙ₊₁−aₙ
C lim aₙ/n
D lim aₙ·n
Root test evaluates L = lim sup √[n]{|aₙ|}. If L<1, the series converges absolutely; if L>1, it diverges. It is useful when ratio test is hard.
If ∑ aₙ converges absolutely, then any rearrangement
A Changes sum always
B Makes it diverge
C Keeps same sum
D Makes it alternating
Absolute convergence guarantees rearrangement invariance: reordering terms does not change the sum. This is not true for conditionally convergent series, where rearrangements can alter sums.
If fₙ are differentiable and fₙ′ converge uniformly with one point fixed, then
A fₙ always diverge
B fₙ converge uniformly
C f is discontinuous
D Only pointwise holds
A standard result: if fₙ′ converge uniformly on an interval and fₙ(x₀) converges at one point x₀, then fₙ converge uniformly to a differentiable function f with f′ as the uniform limit.