hcju/proofwikiqa huggingface dataset:
=== Necessary Condition === Let $\map f x \to l$ as $x \to c$. Then from the definition of the limit of a function: :$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$ So for any given $\epsilon$, there exists a $\delta$ such that: :$0 < \size {x - c} < \delta$ implies that: :$l - \epsilon < \map f x < l + \epsilon$ Now: {{begin-eqn}} {{eqn | o =        | r = 0 < \size {x - c} < \delta       | c =  }} {{eqn | ll= \leadsto       | o =        | r = - \delta < -\paren {x - c} < 0       | c =  }} {{eqn | l = \lor       | o =        | r = 0 < \paren {x - c} < \delta       | c =  }} {{eqn | ll= \leadsto       | o =        | r = c - \delta < x < c       | c =  }} {{eqn | l = \lor       | o =        | r = c < x < c + \delta       | c =  }} {{end-eqn}} That is: $\forall \epsilon > 0: \exists \delta > 0$: :$(1): \quad c - \delta < x < c \implies \norm {\map f x - l} < \epsilon$ :$(2): \quad c < x < c + \delta \implies \norm {\map f x - l} < \epsilon$ So given that particular value of $\epsilon$, we can find a value of $\delta$ such that the conditions for both: :$(1): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the left and : :$(2): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the right. Thus: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ implies that: :$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$ and: :$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$ {{qed|lemma}} === Sufficient Condition === Let $\map f x \to l$ as $x \to c^-$ and $\map f x \to l$ as $x \to c^+$. This means that: :$(1): \quad \forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \size {\map f x - l} < \epsilon$ and : :$(2): \quad \forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \size {\map f x - l} < \epsilon$ In the same manner as above, the conditions on $\delta$ give us that: {{begin-eqn}} {{eqn | o =        | r = c - \delta < x < c       | c =  }} {{eqn | l = \land       | o =        | r = c < x < c + \delta       | c =  }} {{eqn | ll= \leadsto       | o =        | r = 0 < \size {x - c} < \delta       | c =  }} {{end-eqn}} So: :$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$ Thus: :$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$ and: :$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$ together imply that: :$\displaystyle \lim_{x \mathop \to c} \map f x = l$ Hence the result. {{qed}}
=== Necessary Condition === Suppose that: : $\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$ Let $\epsilon > 0$. Then by the definition of the limit of a function: : $\exists \delta > 0: d_2 \left({f \left({x}\right), l}\right) < \epsilon$ provided $0 < d_1 \left({x, c}\right) < \delta$. Now suppose that $\left \langle {x_n} \right \rangle$ is a sequence of points of $S$ such that such that: : $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$ Since $\delta > 0$, from the definition of the limit of a sequence: : $\exists N: \forall n > N: d_1 \left({x_n, c}\right) < \delta$ But: : $\forall n \in \N_{>0}: x_n \ne c$ That means: : $0 < d_1 \left({x_n, c}\right) < \delta$ by the definition of a metric. But that implies that: : $d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$ That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that: : $\forall n > N: d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$ Thus: : $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$ {{qed|lemma}} === Sufficient Condition === Now suppose that for each sequence $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\displaystyle \forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, it is true that: : $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$ What we will try to do is assume that it is ''not'' true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, and try to find a contradiction. So, if it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, then: :$\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < d_1 \left({x, c}\right) < \delta \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon$ where $\land$ denotes logical and. For all $n \in \N_{>0}$, define: :$S_n = \left\{{x \in S: 0 < d_1 \left({x, c}\right) < \dfrac 1 n \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon}\right\}$ By hypothesis, $S_n$ is non-empty for all $n \in \N_{>0}$. Using the axiom of countable choice, there exists a sequence $\left \langle {x_n} \right \rangle$ of points in $S$ such that $x_n \in S_n$ for all $n \in \N_{>0}$. Then: : $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$ but it is ''not'' true that: : $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$ So there is our contradiction, and so the result follows. {{qed}} {{ACC}}
From Linear Function is Continuous, setting $\alpha = 1$ and $\beta = 0$, we have that: :$\displaystyle \lim_{x \mathop \to c} x = c$ Repeated application of the Product Rule for Limits of Functions shows us that: :$\displaystyle \forall k \in \N: \lim_{x \mathop \to c} x^k = c^k$ Now let $\map P x = a_n x^N + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$. By repeated application of the Combined Sum Rule for Limits of Functions, we find that: :$\displaystyle \lim_{x \mathop \to c} \map P x = \map P c$ So whatever value we choose for $c$, we have that $\map P x$ is continuous at $c$. From the definition of continuity on an interval, the second assertion follows. {{qed}}
Let $c \in \R$.  Let $\sequence {x_n}$ be a real sequence converging to $c$.  Then: {{begin-eqn}} {{eqn	| l = \lim_{n \mathop \to \infty} \map f {x_n} 	| r = \lim_{n \mathop \to \infty} \paren {\alpha x_n + \beta} }} {{eqn	| r = \alpha c + \beta 	| c = Combination Theorem for Sequences: Real: Combined Sum Rule }} {{eqn	| r = \map f c }} {{end-eqn}} We therefore have:  :for all real sequences $\sequence {x_n}$ converging to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$. So by Sequential Continuity is Equivalent to Continuity in the Reals $f$ is continuous at $c$.  As $c \in \R$ was arbitrary, $f$ is continuous on $\R$. {{qed}}
Let: :$\map R x = \dfrac {\map P x} {\map Q x}$ be a real rational function, defined at all points of $\R$ at which $\map Q x \ne 0$. Let $c \in \R$. From Real Polynomial Function is Continuous: :$\displaystyle \lim_{x \mathop \to c} \map P x = \map P c$ and: :$\displaystyle \lim_{x \mathop \to c} \map Q x = \map Q c$ Thus by Quotient Rule for Limits of Functions: :$\displaystyle \lim_{x \mathop \to c} \map R x = \lim_{x \mathop \to c} \frac {\map P x} {\map Q x} = \frac {\map P c} {\map Q c}$ whenever $\map Q c \ne 0$. So whatever value we choose for $c$ such that $\map Q c \ne 0$, we have that $\map P x$ is continuous at $c$. From the definition of continuity on an interval, the second assertion follows. {{qed}}
Let $\epsilon > 0$. Since $\displaystyle \lim_{y \mathop \to \eta} \map f y = l$, we can find $\Delta > 0$ such that: :$\size {\map f y - l} < \epsilon$ provided $0 < \size {y - \eta} < \Delta$ Let $y = \map g x$. Then, provided that $0 < \size {\map g x - \eta} < \Delta$, we have: :$\size {\map f {\map g x} - l} < \epsilon$ But $\displaystyle \lim_{x \mathop \to \xi} \map g x = \eta$ and $\Delta > 0$. Hence: :$\exists \delta > 0: \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$ We now need to establish the reason for the conditions under which $0 < \size {x - \xi} < \delta \implies \size {\map f {\map g x} - l} < \epsilon$. As it stands, this is not generally the case, as follows. Consider the functions: :$\map g x = \eta, \map f y = \begin {cases} y_1 & : y = \eta \\ y_2 & : y \ne \eta \end{cases}$ Then: :$\displaystyle \lim_{y \mathop \to \eta} \map f y = y_2$ and $\lim_{x \mathop \to \xi} \map g x = \eta$ But it is not true that $\displaystyle \lim_{x \mathop \to \xi} \map f {\map g x} = y_2$ because $\forall x: \map f {\map g x} = y_1$. Now, if '''Hypothesis 1:''' $f$ is continuous at $\eta$, then $l = \map f \eta$ and so $\size {\map f y - l} < \epsilon$ even when $y = \eta$. So we can write: provided that $\size {\map g x - \eta} < \Delta$, we have: :$\size {\map f {\map g x} - l} < \epsilon$ and the argument holds. Otherwise, let us assume '''Hypothesis 2:''' For some open interval $I$ containing $\xi$, it is true that $\map g x \ne \eta$ for any $x \in I$ except possibly $x = \xi$. Then we can be sure that $\map g x \ne \eta$ provided that $0 < \size {x - \xi} < \delta$ for sufficiently small $\delta > 0$. So we can write: $0 < \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$, and the argument holds. {{Qed}}
Let $x \in \mathbb I$ such that $x$ is not an end point. Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ is the same as $\size {x - y} < \delta$ provided $\delta$ is small enough. The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x} \map f y = \map f x$, that is, that $f$ is continuous at $x$. Now suppose $x \in \mathbb I$ and $x$ is a left hand end point of $\mathbb I$. Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ reduces to $x \le y < x + \delta$ provided $\delta$ is small enough. The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x^+} \map f y = \map f x$, that is, that $f$ is continuous on the right at $x$. Similarly, if $x \in \mathbb I$ and $x$ is a left hand end point of $\mathbb I$, then the criterion reduces to the statement that $f$ is continuous on the left at $x$. Thus the assertions are equivalent to the statement that $f$ is continuous at all points in $\mathbb I$, that is, that $f$ is continuous on $\mathbb I$. {{qed}}
From Limit of Function by Convergent Sequences, we have: :$\displaystyle \lim_{x \mathop \to a} \map f x = \map f a$ {{iff}}: :for each sequence $\sequence {x_n}$ of points of $A_1$ such that: ::$\forall n \in \N_{>0}: x_n \ne a$ :and: ::$\displaystyle \lim_{n \mathop \to \infty} x_n = a$ :it is true that: ::$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$ The result follows directly from this and the definition of continuity. {{qed}}
Let $J$ be the image of $f$. By definition of real interval, it suffices to show that: :$\forall y_1, y_2 \in J: \forall \lambda \in \R: y_1 \le \lambda \le y_2 \implies \lambda \in J$ So suppose $y_1, y_2 \in J$, and suppose $\lambda \in \R$ is such that $y_1 \le \lambda \le y_2$. Consider these subsets of $I$: : $S = \left\{{x \in I: f \left({x}\right) \le \lambda}\right\}$ : $T = \left\{{x \in I: f \left({x}\right) \ge \lambda}\right\}$ As $y_1 \in S$ and $y_2 \in T$, it follows that $S$ and $T$ are both non-empty. Also, $I = S \cup T$. So from Interval Divided into Subsets, a point in one subset is at zero distance from the other. So, suppose that $s \in S$ is at zero distance from $T$. From Limit of Sequence to Zero Distance Point, we can find a sequence $\left \langle {t_n} \right \rangle$ in $T$ such that $\displaystyle \lim_{n \to \infty} t_n = s$. Since $f$ is continuous on $I$, it follows from Limit of Image of Sequence that $\displaystyle \lim_{n \to \infty} f \left({t_n}\right) = f \left({s}\right)$. But $\forall n \in \N_{> 0}: f \left({t_n}\right) \ge \lambda$. Therefore by Lower and Upper Bounds for Sequences, $f \left({s}\right) \ge \lambda$. We already have that $f \left({s}\right) \le \lambda$. Therefore $f \left({s}\right) = \lambda$ and so $\lambda \in J$. A similar argument applies if a point of $T$ is at zero distance from $S$. {{qed}}
As before, let $J$ be the image of $f$. By Subset of Real Numbers is Interval iff Connected we need to show that $J$ is connected (and hence an interval). Suppose not. Then there exists a separation $S \mid T$ of $J$. Define $A = f^{-1}(S)$ and $B = f^{-1}(T)$. $A$ and $B$ are both non-empty. Because $f$ is continuous, by Continuous iff inverse image of any open set is open we must have $A$ and $B$ open. Now, $A \cap B = f^{-1}(S) \cap f^{-1}(T) = f^{-1}(S \cap T) = \varnothing$, because $S \mid T$ is a separation. Also, $A \cup B = f^{-1}(S) \cup f^{-1}(T) = f^{-1}(S \cup T) = f^{-1}(J) = I$ ($S \mid T$ is a separation of $J$). Hence $A \mid B$ is a separation of $I$. $I$ can certainly not be an interval (because it is not connected). This is a contradiction. Thus $J$ must be an interval. {{qed}}
$\mathbb I \subseteq S \cup T \implies \forall x \in \mathbb I: x \in S \lor x \in T$ from the definition of union. That is, every element of $\mathbb I$ belongs either to $S$ or to $T$. The distance of an element $c \in \R$ from a subset $S$ of $\R$ is given as: :$\displaystyle \map d {c, S} = \map {\inf_{x \mathop \in S} } {\size {c - x} }$ Assume that $S$ and $T$ have no element in common, otherwise the result is trivial. {{WLOG}}, suppose that $\exists s \in S, t \in T$ such that $s < t$. (If not, then $\exists s \in S, t \in T$ such that $s > t$, and the following argument may be amended appropriately.) Let $T_0 = \set {x: x \in T: x > s}$. As $t \in T_0$ it follows that $T_0 \ne \O$. Also, $T_0$ is bounded below by $s$. Let $b = \map \inf {T_0}$. If $b \notin T$ then $b \in S$. But from Distance from Subset of Real Numbers, it follows that $\map d {b, T_0} = 0$. Thus we have found an element in $S$ which is zero distance from $T$. Otherwise, $b \in T$. Then $b > s$, and the open interval $\openint s b$ is a non-empty subset of $S$. Hence $b$ is an element in $T$ which is zero distance from $S$. {{Qed}}
From the definition of distance: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$ Consider the set $T = \set {\size {x - y}: y \in S}$. This has $0$ as a lower bound as Absolute Value is Bounded Below by Zero. So: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} } \ge 0$ If $x \in S$ then: :$\size {x - x} = 0 \in T$ and so: :$\displaystyle 0 \le \map {\inf_{y \mathop \in S} } {\map d {x, y} }$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\map d {x, y} } = 0$ {{Qed}}
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space. The result is then seen to be an example of Distance from Subset to Element. {{Qed}}
From the definition of distance: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$ Let $\xi = \sup S$. Then: :$\forall y \in S: \size {\xi - y} = \xi - y$ So we need to show that no $h > 0$ can be a lower bound for $T = \set {\size {\xi - y}: y \in S}$. {{AimForCont}} $\exists h > 0: \forall y \in S: \xi - y \ge h$. But then: :$\forall y \in S: y \le \xi - h$ and hence $\xi - h$ is an upper bound for $T$ smaller than $\xi = \sup S$. But this contradicts the definition of supremum, that is the ''smallest'' upper bound. So there is no such $h > 0$. Hence by Proof by Contradiction it follows that: :$\map d {\xi, S} = 0$. {{Qed}}
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space. The result is then seen to be an example of Distance from Subset to Supremum. {{Qed}}
From the definition of distance: :$\forall x, y \in \R: \map d {x, y} = \size {x - y}$ Thus: :$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$ Let $\xi = \inf S$. Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$. By Negative of Infimum is Supremum of Negatives: :$\xi = \inf S \implies -\xi = \sup S'$ Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result. {{Qed}}
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space. The result is then seen to be an example of Distance from Subset to Infimum. {{Qed}}
First it is shown that: :$\forall n \in \N_{>0}: \exists x_n \in S: \size {\xi - x_n} < \dfrac 1 n$ {{AimForCont}} that: :$\exists n \in \N_{>0}: \not \exists x \in S: \size {\xi - x} < \dfrac 1 n$ Then $\dfrac 1 n$ is a lower bound of the set $T = \set {\size {\xi - x}: x \in S}$. This contradicts the assertion that $\map d {\xi, S} = 0$. We have from Sequence of Powers of Reciprocals is Null Sequence that: :$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 n = 0$ So as $\size {\xi - x_n} < \dfrac 1 n$ it follows from the Squeeze Theorem for Real Sequences that: :$\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$ Hence the result. {{qed}}
Suppose $f$ is not bounded on $\closedint a b$. Then from the corollary to Limit of Sequence to Zero Distance Point, there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$. Since $\closedint a b$ is a closed interval, from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_n}$ which converges to some $\xi \in \closedint a b$. Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$. But this contradicts our supposition that there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$. The result follows. {{qed}}
Let $\sequence {x_n}$ be a sequence in $\closedint a b$. Since $\closedint a b$ is bounded in $\R$, it follows that $\sequence {x_n}$ is a bounded sequence. By the Bolzano-Weierstrass Theorem, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which is convergent. Suppose $x_{n_r} \to l$ as $n \to \infty$. Since $a \le x_{n_r} \le b$, from Lower and Upper Bounds for Sequences it follows that $a \le l \le b$. So $\sequence {x_{n_r} }$ converges to a point in $\closedint a b$. {{qed}}
From Image of Closed Real Interval is Bounded, we have that $f$ is bounded on $\closedint a b$. Let $d$ be the supremum of $f$ on $\closedint a b$. Consider a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to d$ as $n \to \infty$. From the corollary to Limit of Sequence to Zero Distance Point, this can always be found. Now $\closedint a b$ is a closed interval So from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which converges to some $\xi \in \closedint a b$. Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$. So $\map f \xi = d$ and thus the supremum $d$ is indeed a maximum. A similar argument shows that the infimum is a minimum. {{qed}}
This is an instance of the Extreme Value Theorem. $\closedint a b$ is a compact subset of a metric space from Real Number Line is Metric Space. $\R$ itself is a normed vector space. {{MissingLinks|Find the reference to $\R$ being a normed vector space.}} Hence the result. {{qed}}
Let $I = \left[{a \,.\,.\, b}\right]$. Let $J = f \left({I}\right)$. From Image of Interval by Continuous Function is Interval, $J$ is an interval. From Image of Closed Real Interval is Bounded, $J$ is bounded. From Max and Min of Function on Closed Real Interval, $J$ includes its end points. Hence the result. {{qed}}
{{MissingLinks|one-dimensional, for a start}} Aiming for a contradiction, suppose such a smooth mapping exists. By the Morse-Sard Theorem, there exists a regular value $x \in \partial M$. By the Preimage Theorem: :$\map {f^{-1} } x$ is a submanifold of $M$ with boundary. We have that the codimension of $\map {f^{-1} } x$ in $M$ equals the codimension of $x$ in $\partial M$, that is, $\map \dim M - 1$. Then $\map {f^{-1} } x$ is one dimensional and compact. Since $\partial f$ is the identity mapping: :$\map {\partial f^{-1} } x = \map {f^{-1} } x \cap \partial M = \set x$ This contradicts the Classification of Compact One-Manifolds. {{qed}} Category:Algebraic Topology Category:Named Theorems 5bhlxb7envkzp480dmd6wk5jwgarz0w
=== Lemma 1 === Let $f$ be a function on $\closedint a b$ that is smooth and has a positive derivative everywhere except one interior point, $c$. Then there exists a globally smooth function $g$ that agrees with $f$ near $a$ and $b$ and has a positive derivative everywhere. ==== Proof of Lemma 1 ==== Let $r$ be a smooth nonnegative function that vanishes outside a compact subset of $\openint a b$, which equals $1$ near $c$, and which satisfies $\displaystyle \int_a^b r = 1$. Define: :$\displaystyle \map g x = \map f a + \int_a^x \paren {k \map r s + \map {f'} s \paren {1 - \map r s} } \rd s$ where the constant $\displaystyle k = \map f b= - \map f a - \int_a^b \map {f'} s \paren {1 - \map r s} \d s$. {{explain|why this function g satisfies the given criteria}} {{qed|lemma}} Let $f$ be a Morse function on a one-manifold $X$. Let $S$ be the union of the critical points of $f$ and $\partial X$. As $S$ is finite, $X - S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$. === Lemma 2 === $f$ maps each $L_i$ diffeomorphically onto an open interval in $\R$ ==== Proof of Lemma 2 ==== Let $L$ be any of the $L_i$. Because $f$ is a local diffeomorphism and $L$ is connected, $f \sqbrk L$ is open and connected in $\R$. We also have $f \sqbrk L \in f \sqbrk X$, the latter of which is compact. Hence there are numbers $c$ and $d$ such that $f \sqbrk L = \openint c d$. It suffices to show $f$ is one to one on $L$, because then $f^{-1}: \openint c d \to L$ is defined and locally smooth. Let $p$ be any point of $L$. Set $q = \map f p$. It suffices to show that every other point $z \in L$ can be joined to $p$ by a curve $\gamma: \closedint q y \to L$ such that $f \circ \gamma$ is the identity and $\map \gamma y = z$. Since $\map f z = y \ne q = \map f p$, this result shows $f$ is one to one. So let $Q$ be the set of points $x$ that can be so joined. Since $f$ is a local diffeomorphism, $Q$ is both open and Definition:Closed Set (Topology). Hence $Q = L$. {{qed|lemma}} === Lemma 3 === Let $L$ be a subset of $X$ diffeomorphic to an open interval in $\R$, where $\dim X = 1$. Then the closure $\map \cl L$ contains at most two points not in $L$. ==== Proof of Lemma 3 ==== Let $g$ be a diffeomorphism $g: \openint a b \to L$ and let $p \in \map \cl L - L$. Let $J$ be a closed subset of $X$ diffeomorphic to $\closedint 0 1$ such that: :$1$ corresponds to $p$ :$0$ corresponds to some $\map g t$ in $L$. Consider the set $\set {s \in \openint a t: \map g s \in J}$. This set is both open and closed in $\openint a b$. Hence $J$ contains either $g \sqbrk {\openint a t}$ or $g \sqbrk {\openint t b}$. {{qed|lemma}} {{finish}} === Proof of Corollary === Follows trivially. {{qed}} {{proofread}} Category:Topology r8ydxytmtr0gkww1oq5cd0c3xh3w2f3
By hypothesis, $\map {f'} {x_0}$ exists. We have: {{begin-eqn}} {{eqn | l = \map f x - \map f {x_0}       | r = \frac {\map f x - \map f {x_0} } {x - x_0} \cdot \paren {x - x_0}       | c =  }} {{eqn | o = \to       | r = \map {f'} {x_0} \cdot 0       | c = as $x \to x_0$ }} {{end-eqn}} Thus: :$\map f x \to \map f {x_0}$ as $x \to x_0$ or in other words: :$\displaystyle \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$ The result follows by definition of continuous. {{qed}}