MathematicalReal Analysis
Question28
a)f: X → Y {displaystyle fcolon X
ightarrow Y} Iff is open, it must not necessarily be continuous.
Proof:
“Forthe function f : M→N between two topological spaces Mand Nis continuous if for every open set V⊆N,the inverse image f− 1 ( V ) = { x ∈X  f ( x ) ∈V } {displaystyle f^{1}(V)={xin Xf(x)in V}} f^{1}(V)= { m⊆M f(m) ⊆ V} is an open subset of M.That is, fis a function between the sets Mand N(not on the elements of the topology T_{M}),but the continuity of fdepends on the topologies used on Mand N.”
“Anextreme example: if a set Xis given the discrete topology (in which every subset is open), allfunctionsofany topological space Tare continuous. On the other hand, if Mis equipped with the indiscrete topology (in which the only opensubsets are the empty set and M)and the space Tset is at least T_{0},then the only continuous functions are the constant functions.Conversely, any function whose range is indiscrete is continuous.”
B)If f is a homeomorphism, it is open. By definition, ahomeomorphism is a continuous function with a continuous inversefunction.
Proof:“If a function f: M→N is between two topological spaces (M, T_{M})and (N, T_{N}),then it is called a homeomorphism if the following hold:”

f is a bijection

f is continuous

The inverse function f^{1} is continuous.
Givennow that f is a homeomorphism, it means it has an inverse function,and if it has an inverse function, then it is automatic f is open.
Therefore,if f is a homeomorphism, then it is open.
c) If f is an open, continuous bijection, then it is a homeomorphism.
Thesame conditions that hold for a function to be a homeomorphism stillapply here,
Fmust be a bijection, continuous and with the inverse function.
d)if f: R→R is a continuous surjection, it is not necessarily that fmust be open.Proof:Consider f: R → R defined by f(x) = x for x ≤ 0,f(x) = 0 for 0 ≤ x ≤ 1, and f(x) = x – 1 for 1 ≤ x.f is a continuous surjection but (0,1) is open while f(0,1) ={0} is not. → f is not an open mape)fis homeomorphism if it is a continuous, open bijection, so itsuffices to show that f is injective. Supposefor contradiction that f(s) = f(t) for s < t. [s,t]is compact and connected so it`s image is compact and connected,meaning f[s,t] = [m,M] for some m ≤ M. Iff(s) = f(t) = m then f(s,t) = (m,M] or [m,M], depending on whether ornot there exists u ∈(s,t) with f(u) = m. Similarly,if f(s) = f(t) = M then f(s,t) = [m,M) or [m,M]. If m < f(s) =f(t) < M then by connectedness of (s,t) we have f(s,t) = [m,M]. Inany of these cases, (s,t) is open but f(s,t) is not, contradictingopenness of f. → f is injective and therefore a homeomorphism.f)We identify S^{1}with the set of complex numbers of unit length. Themap defined by x ↦x² is a continuous, open surjection, butnot injective, and therefore not a homeomorphism.
Question44
a)If f is continuous, then its graph is closed
Proof:
Here,we seek to show theboundedness.Sowe`re given that fisbounded and its graph, which we can call G,is a closed subset of R2.
Toshow that it`s continuous, we can show any one of a list ofequivalent conditions, including the inverse image of an open set isopen, or the inverse image of a closed set is closed. But f−1(S),for S⊂R,is just the projection of (R×S)∩Gonto the xaxis.
Ifyou utilize the fact that G⊂R×I,where Iis a sufficiently large closed interval, and consider f−1(C),Ca closed subset of I,I believe compactness is on your side to show that f−1(C)contains all of its limit points.
B)If f is continuous and M is compact, then its graph is compact.
Proof:
Iff is bounded let Mbe a compact interval with f[X]⊂M.The graph Γ={(x,f(x))∣x∈R}of fis a subset of R×Mby assumption. If now Kis a closed subset of M,then note that we have
f−1[K]=π[Γ∩(R×K)](as x∈f−1[K] then f(x)∈Kand (x,f(x))is in Γ and in R×Kand xis its image under π,while if x=π(x,y)for y∈Γ∩R×Kwe know y∈Kand y=f(x)to be on Γ, so x∈f−1[K])
c) If the graph of f is compact, then f is continuous.
Proof:
Assumethat the graph of fis compact. It means that it is also closed and bounded. The graph isa closed and bounded subset of A×f(A).All we need to show is that f(A)is compact.
IfA is a compact metric space, for all functions f: A→R, f(A) compactimplies fcontinuous.
Thisclaim is false. For example, consider A=[−1,1]
asa metric subspace of R and define f:A→Rby
f(x)={_{1otherwise}^{1if x˃o}
Thusf(A)is the compact set {−1,1} but fis not continuous.
d)Hereis an example of discontinuous function: f(x)=3^{x/1x2}
Thegiven function is not defined at x=−1and x=1.Hence, this function has discontinuities at x=±1.To determine the type of the discontinuities, we find the onesidedlimits.
Question55
a)We understand that a limit is a value that a function or a sequenceapproaches as the index approaches some value. In this case, if p isa limit of S, then the dist (p, S) = 0
Proof:
Ifp is a limit point, then for every e there is an N such that n> N ==> p – 1/n < e Ifp is negative, then for all positive n, p – 1/n > p so we havea contradiction. Ifp >= 1, then set e = 1/2. For all n > 2, p – 1/n > 1/2, soagain we have a contradiction. Ifp < 1 and greater than 0, let P = 1/p, let K be the first integergreater than P, and let e = (1/K – 1/(K+1)). Then you can show thatfor all n > K+1, p – 1/n > e.

If dist(p,s) = inf{d(p,s) : s с S } then dist( p,s) = 0 , therefore, using the triangle inequality, we have that dist(p,s) ≤ dist ( p,s) + dist (s,S) and it stands that p→dist(p,s) is a continuous function of p с M. That is, with a single space (M,d) , d: M X M → R is uniformly continuous.