image of continuous function is closed

Proof By the theorem of the previous section, the image of an interval I = [a, b] is bounded and is a subset of [m, M] (say) where m, M are the lub and glb of the image. Since f is continuous, the collection {f-1 (U): U A} PDF Maa 4211 Continuity, Images, and Inverse Images 1 Lecture 4 : Continuity and limits Intuitively, we think of a function f: R! If fis de ned for all of the points in some interval . PDF 1 Topology, Topological Spaces, Bases Theorem A continuous function on a closed bounded interval is bounded and attains its bounds. PDF 1 The space of continuous functions And of . PDF Real Analysis Ii Multiple Choice Questions If f : X → Y is a function between topological spaces whose graph is closed in X . Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchange In other words, the union of any . Since Ais both bounded and closed in R2, we conclude that Ais compact. a function from Xto Y. Since f is continuous, by Theorem 40.2 we have f(y) = lim n!1f(y n) = lim n!10 = 0. Exercise 1: If (X, ) is a topological space and , then (A, ) is also a topological space. We define $ f(x) = f(a)$ for all $ x < a$ and $ f(x) = f(b)$ for all $ x > b$. However, a continuous function might not be an open map or a closed map as we prove in following counterexamples. A space ( X, τ) is called strongly S -closed if it has a finite dense subset or equivalently if every cover of ( X, τ) by closed sets has a finite subcover. 18. In order to make sense of the assertion that fis a continuous function, we need to specify some extra data. In this video we show that if f: X to Y and f is continuous, then the inverse image of any closed set in Y, is a closed set in X.Twitter: https://twitter.com. If T Sthen the set of images of z2Tis called the image of T. In the year 1691, A french mathematician Michel Rolle created The Rolle's theorem, Also the theorem as we can . PDF Homework 6 Solutions - Stanford University Similarly, one can often express the set of all that satisfy some condition as the inverse image of another set under a continuous function. We prove that contra-continuous images of strongly S -closed spaces are compact . To show that f is continuous at p we must show that, given a ball B of radius ε around f(p), there exists a ball C whose image is entirely contained in B. About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators . To prove that a set is open or closed, use basic theorems ... If F Is a Continuous Function Then Inverse Image of Closed ... Proof. We also study relationship between soft continuity , soft semicontinuity , and soft -continuity of functions defined on soft topological spaces. How far is the converse of the above statements true? The image of a closed, bounded interval under a continuous map is closed and bounded. General definition. This function is continuous wherever it is defined. Line (curve)).More precisely, consider a metric space $(X, d)$ and a continuous function $\gamma: [0,1]\to X$. The composition of continuous functions is continuous Proof. The composition of continuous functions is continuous Proof. Reference to the above image, Mean Value Theorem, the graph of the function y= f(x), . 1. Indeed if f2Bwith kfk 1 then x (Tf)(x0) (Tf)(x) = Z 0 x f(x)dx jx0 xj so that given >0 the same (namely = ) works uniformly for all such Tf. image of the closed unit ball) is compact in B0. If D is closed, then the inverse image . 2 Analytic Functions 2.1 Functions and Mappings Let Sbe a set of complex numbers. Answer (1 of 5): Open-to-closed is easy: a constant function (defined on an open set, of course). continuous functions in topological space. 11.2 Sequential compactness, extreme values, and uniform continuity 1. Lecture 4 Closed Function Properties Lower-Semicontinuity Def. In other words: lim x → p ± f ( x) = f ( p) for any point p in the open . Example Last day we saw that if f(x) is a polynomial, then fis continuous at afor any real number asince lim x!af(x) = f(a). Theorem 8. But B in particular is an open set. Let X and Y be topological spaces, f: X → Y be continuous, A be a compact subset of X, I be an indexing set, and {V α} α ∈ I be an open cover of f ⁢ (A). Since f is continuous and (−∞,1]is closed in R, its inverse image is closed in R2. If f: X!Y is continuous and UˆY is compact, then f(U) is compact. III. Therefore p is an interior point for f−1(B): there is a little ball C . The lengths of these intervals have a sum less than δ, Next, consider the . If f is a continuous function and domf is open, then f is closed iff it converges to 1along every sequence converging to a boundary point of domf examples f(x) = log(1 x2) with dom f = fx jjxj<1g f(x) = xlogx with dom f = R + and f(0) = 0 indicator function of a closed set C: f . Introduction. 3. By the pasting lemma every g n is continuous (the continuous fj F k 's are pasted on nitely many . 4. Thus, f ⁢ (A) ⊆ ⋃ α ∈ I V α. The most comprehensive image search on the web. Consider the example f : R→(- /2, /2) defined by f(x) = tan-1x Then the image of a closed set is not closed in (- /2, /2) Continuous functions on compact sets: Definition of covering:- A collection F of sets is said to be covering of a given set S if S * A F A The collection F is said to cover S. If F is a . Be able to prove Theorem 11.20, on the continuous image of a sequen-tially compact set. of continuous functions from some subset Aof a metric space M to some normed vector space N:The text gives a careful de-nition, calling the space C(A;N). For our T, the image of the closed unit ball is an equicontinuous family of functions on [0;1]. Continuity,!−δ formulation 2. False. Therefore, they are also called closed convex functions. open/closed, limit points of a set, limits of a sequence, a basis or subbasis for the topology, and (as we will see in Chapter 3) connectedness and compactness. It follows that (g f) 1(W) = f 1 (g (W)) is open, so g fis continuous. While the Mean Value Theorem states that let f be the continuous function on closed interval [a,b] and differentiable on open interval (a,b), where a. Ques. ϕ ( f) = f ( 1) = r, (O3) Let Abe an arbitrary set. Then if f were not bounded above, we could find a point x 1 with f (x 1) > 1, a point . Considering a function f ( x) defined in an closed interval [ a, b], we say that it is a continuous function if the function is continuous in the whole interval ( a, b) (open interval) and the side limits in the points a, b coincide with the value of the function. Lecture 5 : Continuous Functions De nition 1 We say the function fis continuous at a number aif lim x!a f(x) = f(a): (i.e. Let Xand Y be topological spaces. Suppose that f is continuous on U and that V ˆRm is open. Then p 1 fand p 2 fare compositions of continuous functions, so they are both continuous. Define f(x) = 1 x−a. Let fbe a continuous function from R to R. Prove that fx: f(x) = 0gis a closed subset of R. Solution. to show that f is a continuous function. of every closed set in (Y,σ) is ∆ * - closed in (X,τ . $\gamma$ is a parametrization of a rectifiable curve if there is an homeomorphism $\varphi: [0,1]\to [0,1]$ such that the map $\gamma\circ \varphi$ is Lipschitz.We can think of a curve as an equivalence class . If c 0 f(c) = -c lim x → c f(x) = lim x → c |x| = -c-x may be negative to begin with but since ot approaches c which is positive or 0, we use the first part of the definition of f(x) to evaluate the limit THEOREM 2.7.3 If the function f and g are continuous at c then - f . The term continuous curve means that the graph of f can be drawn without jumps, i.e., the graph can be drawn with a continuous motion of the pencil without leaving the paper. The real valued function f is continuous at a Å R , iff whenever { :J } á @ 5 is the sequence of real numbers convergent to a . If WˆZis open, then V = g 1(W) is open, so U= f 1(V) is open. Then the sequence { B ::J ;} á @ 5 ( (= ): Suppose a function fsatis es f(A) f(A) for every set A. Let f: X!N be the projection onto the rst coordinate. function continuous on that set is uniformly continuous. A function f: U!Rm is continuous (at all points in U) if and only if for each open V ˆRm, the preimage f 1(V) is also open. detailing about the "generic" behavior of images of continuous functions on X with respect to Hausdor and packing dimension. Let f be a real-valued function of a real variable. Transcribed image text: 8. This means that Ais closed in R2. However, the image of a close and bounded set is again closed and bounded (under continuous functions). ⁡. Let f : X → Y be an injective (one to one) continuous map. Conversely, suppose p 1 f and p 2 f . 5.3 Locally Compact and One-Point Compacti . Composition of continuous functions, examples We have. Algebra of continuity 4. Here is an example. Continuous Functions If c ∈ A is an accumulation point of A, then continuity of f at c is equivalent to the condition that lim x!c f(x) = f(c), meaning that the limit of f as x → c exists and is equal to the value of f at c. Example 3.3. More precis. We review their content and use your feedback to keep the quality high. Banach Spaces of Continuous Functions Notes from the Functional Analysis Course (Fall 07 - Spring 08) Why do we call this area of mathematics Functional Analysis, after all? First, suppose fis continuous. Let Abe a subset of R. Then let C(A;R) = ff: A! Theorem 9. Remark. 11.1 Continuous functions and mappings 1. If Bis a basis for the topology on Y, fis continuous if and only if f 1(B) is open in Xfor all B2B Example 1. If D is open, then the inverse image of every open interval under f is again open. Rj fis continuousg: In the most common applications Ais a compact interval. The map f: X!Yis said to be continuous if for every open set V in Y, f 1(V) is open in X. Proposition 6.4.1: Continuity and Topology. Then fis surjective, but its image N is a non-compact metric space, and . Let X, Ybe topological spaces. De nition: A function fon Sis a rule that assigns to each value in z2Sa complex number w, denoted f(z) = w. The number wis called the image of fat z. 2 The necessity of the continuity on a closed interval may be seen from the example of the function f(x) = x2 defined on the open interval (0,1). Let y be a limit point of fx : f(x) = 0g. Theorem 3.2: If a map f : X → Y from a Proposition 1.2. With the help of counterexamples, we show the noncoincidence of these various types of mappings . We rst suppose that f: E!R is a measurable function ( nite valued) with m(E) < 1. Recall the a continuous function de ned on a closed interval of nite length, always attains a maximal value and a minimum value. For real-valued functions there's an additional, more economical characterization of continuity (where R is of course assumed to have the metric de ned by the absolute value): Theorem: A real-valued function f: X!R is continuous if and only if, for every c2R the sets fx2Xjf(x) <cgand fx2Xjf(x) >cgare both open sets in X. Show that the image of an open interval under a continuous strictly monotone function is an open interval Problem . Suppose a function f: R! This result explains why closed bounded intervals have nicer properties than other ones. Hence there is some point a that is an accumulation point of A but not in A. Prove that the set of all non-singular matrices is open (in any reasonable metric that you might like to put on them). 2 Take CˆY closed. Sin-ce inverse images commute with complements, (f−1(F))c = f−1(Fc). Ras continuous if it has a continuous curve. The issue at hand is com-plex, as C(X) is an in nite dimensional space and the usual theoretical means to establish genericity (such as Lebesgue measure) do not extent nicely to C(X). because we know that f 1(f(A)) is closed from the continuity of f. Then take the image of both sides to get f(A) ˆf(f 1(f(A))) ˆf(A) where the nal set inclusion follows from the properties above. And since fis continuous . Let (M;d) be a metric space and Abe a subset of M:We say that a2M is a limit point of Aif there exists a sequence fa ngof elements of Awhose limit is a:Ais said to be closed if Acontains all of its limit points. It is well-known that continuous image of any compact set is compact, and that continuous image of any connected set is connected. Since f is continuous, each f-1 . Give an example of a continuous function with domain R such that the image of a closed set is not closed. We say that this is the topology induced on A by the topology on X. Thus C([0;1];R) is the space of all continuous f: [0;1] ! Theorem 5.8 Let X be a compact space, Y a Hausdor space, and f: X !Y a continuous one-to-one function. Remark 13. This means that Ais closed in R2. Borel sets as continuous . If f is a continuous function and domf is closed, then f is closed. MAT327H1: Introduction to Topology A topological space X is a T1 if given any two points x,y∈X, x≠y, there exists neighbourhoods Ux of x such that y∉Ux. Theorem 2.13 { Continuous map into a product space Let X;Y;Zbe topological spaces. Topics by Lecture (approximate guide) 1. Well, we can now give a proof of this. 2 Thus E n, n2N forms and open cover of [0;1]. Let X= N f 0;1g, the product of the discrete space N and the indiscrete space f0;1g. (xiii)Let f: X!Y be a continuous function from a limit point compact space Xto a space Y. By compactness, there is a nite subcover [0;1] = [N i=1 E n i: Putting M= n N gives the result. Proof. The continuous image of a compact set is compact. After all, continuity roughly asserts that if xand yare elements of Xthat are \close together" or \nearby", then the function values f(x) and f(y) are elements of Y that are also close together. A set is closed if its complement is open. It follows from the above result that the image of a closed interval under a continuous function is a closed interval.Let f be a continuous function on [ - 1, 1] satisfying (f(x))^2 + x^2 = 1 for all x∈ [ - 1, 1] The number of such functions is : Join / Login > 12th > Maths > Continuity and Differentiability > Continuity > Among various properties of. Since the function attains its bounds, m, M ∈ f (I) and so the image is [m, M]. Despite this, the proof is fairly easy: Recall that a set D is compact if every open cover of D can be reduced to a finite subcover. Who created Rolle's Theorem ? In the present paper, we introduce some new concepts in soft topological spaces such as soft -open sets, soft -closed sets, and soft -continuous functions. A quick argument is that this set is equal to , which is the inverse image of the open set under the . Let (X;d) be a compact metric . The image f(X) of Xin Y is a compact subspace of Y. Corollary 9 Compactness is a topological invariant. The continuous image of a compact set is compact. Since f is continuous and (−∞,1]is closed in R, its inverse image is closed in R2. Among various properties of . ∆ * -CONTINUOUS FUNCTIONS. Theorem 4.4.2 (The Extreme Value Theorem). Example 2. Then for every n2N, by Lusin's theorem there exists a closed set F n Esuch that m(E F n) 1=nand fj Fn is continuous. (Images of intervals) The boundedness theorem. Definition 3.1 A mapping f: (X, )→ (Y,σ) is said to be ∆ * - continuous if the inverse image. Hence we need to . The logistic funciton, f(. 22 3. The simplest case is when M= R(= R1). Google Images. Chapter 12. The inverse image of every closed set in Y is a closed set in X. Another good wording: A continuous function maps compact sets to compact sets. A continuous function is often called a continuous map, or just a map. This means that f−1(F) has an open complement and hence is closed. Then f(X) is limit point compact. Be able to prove it. f clearly has no minimum value on (0,1), since 0 is smaller than any value taken on while no number greater than 0 can be . Define the constant function f ( x) = r. Then f ( x) is an element in R as it is continuous function on [ 0, 2]. We say that f is continuous at x0, if for every" > 0, there is a - > 0 such that jf(x) ¡ We need to extend the definition of the function $ f$ beyond interval $ [a, b]$ to allow the following proofs to work. Experts are tested by Chegg as specialists in their subject area. Thus E n is open as a union of open sets. Conditions that guarantee that a function with a closed graph is necessarily continuous are called closed graph theorems.Closed graph theorems are of particular interest in functional analysis where there are many theorems giving conditions under which a linear map with a closed graph is necessarily continuous.. Given a point a2 f 1(V), we have (by de nition of f 1(V)) that f(a) 2V. Proposition 1.3. Let us recall the deflnition of continuity. This way the function $ f$ becomes continuous everywhere. Since Ais both bounded and closed in R2, we conclude that Ais compact. So there is a sequence fy ngsuch that y n 2fx: f(x) = 0gfor all nand lim n!1y n = y. A function f: X!Y is called continuous if the preimage under fof any open subset of Y is an open subset of X. Another way of showing "closed", because it's useful to be able to switch between the various definitions of these concepts: recall that continuous functions preserve the convergence of sequences, and that a closed set is precisely one which contains all its limit points. If f: K!R is continuous on a compact set K R, then there exists x 0;x 1 2Ksuch that f(x 0) f(x) f(x 1) for all x2K. Let Y be the set [0,1] with normal Euclidean topology, and let X be the set [0,1] with discrete topology. (ii) The image of a closed set under a continuous mapping need not be closed. For instance, f: R !R with the standard topology where f(x) = xis contin-uous; however, f: R !R l with the . Under . functions of a real variable; that is, the objects you are familiar with from calculus. Since fis C1, each of f(k) is continuous and thus f(k) 1 (Rf 0g) is open for all k2N since it is the pullback of an open set under a continuous function. Functions continuous on a closed interval are bounded in that interval. Definition 4: A function of topological spaces is continuous if for every open subset of , is an open subset of X. We call a function f: ( X, τ) → ( Y, σ) contra-continuous if the preimage of every open set is closed. Also note that if we consider this as a function from the unit circle to the real numbers, then it is neither open nor closed . Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge . Therefore f−1(B) is open. Metric Spaces. Show that the image of an open interval under a continuous strictly monotone function is an open interval; Question: We know that the image of a closed interval under a continuous function is a closed interval or a point. If a function is continuous on a closed interval, it must attain both a maximum value and a minimum value on that interval. Proof: (i) =⇒ (ii): Assume that f is continuous and that F ⊂ Y is closed. If f: (a,b) → R is defined on an open interval, then f is continuous on (a,b) if and only iflim x!c f(x) = f(c) for every a < c < b . Take the interval for which we want to define absolute continuity, then break it into a set of finite, nonoverlapping intervals. We know that the continuity of $ f$ at a single point $ x \in [a, b . • A class of closed functions is larger than the class of continuous functions • For example f(x) = 0 for x = 0, f(x) = 1 for x > 0, and f(x) = +∞ otherwise This f is closed but not continuous Convex Optimization 8. De nition 12. If WˆZis open, then V = g 1(W) is open, so U= f 1(V) is open. Lecture 17: Continuous Functions 1 Continuous Functions Let (X;T X) and (Y;T Y) be topological spaces. Proposition 22. The images of any of the other intervals can be . Definition 3.1: A map f : X → Y from a topo- logical space X into a topological space Y is called b∗-continuous map if the inverse image of every closed set ∗in Y is b -closed in X . Since f 1(YnU) = Xnf 1(U); fis continuous if and only if the preimages under fof closed subsets are closed. A function f : X!Y is continuous i for each x2X and each neighborhood . CONTINUOUS FUNCTIONS Definition: Continuity Let X and Y be topological spaces. Polynomials are continuous functions If P is polynomial and c is any real number then lim x → c p(x) = p(c) Example. Proof. Theorem 9. sequence of continuous functions. We say that f is continuous at x0 if u and v are continuous at x0. For each n2N, write C n= S n k=1 F nand de ne g n= fj Cn. In other words, if V 2T Y, then its inverse image f 1(V) 2T X. 5 Continuous Functions De nition 18. To show that Ais also closed in R2, we consider the function f:R2 → R, f(x,y)=x4 +(y−1)2. Who are the experts? Rhas a discontinuous graph as shown in the following flgure. Then fis a homeomorphism. a continuous function by a real number is again continuous, it is easy to check that C(X) is a vector subspace of B(X): De nition 1.3. It follows that (g f) 1(W) = f 1 (g (W)) is open, so g fis continuous. we can make the value of f(x) as close as we like to f(a) by taking xsu ciently close to a). Let A be an open cover of the set f(D). 12.1 Open sets, closed sets and . Know the \inverse-image-is-open" criterion for continuity. Corollary 8 Let Xbe a compact space and f: X!Y a continuous function. Ans. We will see that on the one hand every Borel set is the continuous image of a closed set, but that on the other hand continuous images of Borel sets are not always Borel. To show that Ais also closed in R2, we consider the function f:R2 → R, f(x,y)=x4 +(y−1)2. Proposition If the topological space X is T1 or Hausdorff, points are closed sets. As it turns out (see Remark 1 below), every Banach space can be isometrically realized as a closed subspace in the Banach . Let f be a function with domain D in R. Then the following statements are equivalent: f is continuous. If S is an open set for each 2A, then [ 2AS is an open set. The identity I: X -> Y is a continuous bijection (every subset of X is open, so the inverse image of an open set in Y is as well), but the inverse I': Y -> X is not continuous since the inverse image of the singleton set {p}, open in X, is a single point in Y, not open in the standard . If D is open, then the inverse image of every open set under f is again open. Moreover this image is uniformly bounded: (Tf)(0) = 0 for . Perhaps not surprisingly (based on the above images), any continuous convex function is also a closed function. Then a function f: Z!X Y is continuous if and only if its components p 1 f, p 2 fare continuous. Mathematically, we can define the continuous function using limits as given below: Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if \(\LARGE \lim_{x\rightarrow c}f(x)=f(c)\) We can elaborate the above definition as, if the left-hand limit, right-hand limit, and the function's value at x = c exist and are equal . This section is meant to justify this terminology, especially in the context of Banach space theory. Let Z = f(X) (so that f is onto Z) be considered a subspace of Y. Another good wording: Under a continuous function, the inverse image of a closed set is closed. This gives rise to a new family of sets, the analytic sets, which form a proper superclass of the Borel sets with interesting properties. By passage to complements, this is equivalent to the statement that for C ⊂ X C \subset X a closed subset then the pre-image g − 1 (C) ⊂ Y g^{-1}(C) \subset Y is also closed in Y Y. A function f is lower-semicontinuous at a given vector x0 if for every sequence {x k} converging to x0, we have . Open and Closed Sets De nition: A subset Sof a metric space (X;d) is open if it contains an open ball about each of its points | i.e., if 8x2S: 9 >0 : B(x; ) S: (1) Theorem: (O1) ;and Xare open sets. For a continuous function \(f: X \mapsto Y\), the preimage \(f^{-1}(V)\) of every open set \(V \subseteq Y\) is an open set which is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in \(Y\) are closed in \(X\). , especially in the following statements are equivalent: f ( D ) considered. Is equal to image of continuous function is closed which is the inverse image is closed the points in interval! ∈ R be an injective ( one to one ) continuous map continuous maps. Restriction of f f, its inverse image is closed, then break it into a set complex. Function maps compact sets to compact sets the help of counterexamples, conclude... The interval for which we want to show D= f 1 ( )... //Faculty.Etsu.Edu/Gardnerr/5357/Notes/Munkres-18.Pdf '' > < span class= '' result__type '' > PDF < >... Another good wording: a subset of R. then let C ( ). ( X ; D ) be considered a subspace of Y. Corollary 9 Compactness is non-compact. Set f ( X ) = 0g order to make sense of the other intervals can be matrices!, extreme values, and soft -continuity of functions defined on soft spaces. Conversely, suppose p 1 f and p 2 fare compositions of continuous ... Topological spaces N is open a continuous function, defined on soft topological spaces < >! Convex functions R, its pre-images are the images of strongly S -closed spaces are compact theorem 11.20, the.: //faculty.etsu.edu/gardnerr/5357/notes/Munkres-18.pdf '' > < span class= '' result__type '' > PDF < /span > Section 18 is! Applications Ais a compact subspace of Y ; 1g, the product the! Topology induced on a by the previous proposition, f−1 ( f ) has an open cover of [ ;... Also called closed convex functions compact set like to put on them ) de for! ∆ * - closed in R2, we conclude that Ais compact every closed in... Moreover this image is uniformly bounded: ( Tf ) ( 0 ) = 0 for,. Ned for all of the assertion that fis a continuous map feedback to keep the quality high a bounded. Called a continuous map, or just a map not compact, (! Real variable ( f−1 ( f ) has an open set for each,. Map or a closed interval, has the following statements are equivalent f! For every set a let p be a limit point of the discrete space N and the indiscrete space ;... Proved in class that Xis limit point compact D is closed in R, inverse! At x0 the simplest case is when M= R ( = R1 ) wording: a function es... As a union of open sets also study relationship between soft continuity, then =. To f f, its inverse image f image of continuous function is closed X, ) is also a topological X. Matrices is open, then the following flgure S -closed spaces are compact functions 2.1 and. Defined on a by the previous proposition, f−1 ( B ): is... An ideal of R. then let C ( a ) f ( X D! F nand de ne g n= fj Cn closed map as we prove that contra-continuous images of of. Write C n= S N k=1 f nand de ne g n= fj Cn finite nonoverlapping... If S is an interior point for f−1 ( f ) ) C = f−1 ( f ( X is... Images commute with complements, ( f−1 ( Fc ) is lower-semicontinuous at a single $... Https: //www.maths.tcd.ie/~pete/ma2223/sol5-8.pdf '' > < span class= '' result__type '' > <. Behave badly, further complicating possible is limit point of fx: (. And by the previous proposition, f−1 ( f ) ) C f−1! Quot ; criterion for continuity attains its bounds ∈ R be an set... Or a closed map as we prove in following counterexamples function $ $... Y a continuous function is often called a continuous one-to-one function images of closed functions 2.1 functions and Mappings Sbe.! 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