nyuuzyou/wikis huggingface dataset:
Line Parallel to Side of Triangle which Bisects One Side also Bisects Other Side

== Theorem ==
Let $ABC$ be a triangle.
Let $DE$ be a straight line parallel to $BC$.
Let $DE$ bisect $AB$.
Then $DE$ also bisects $AC$.
That is, $DE$ is a midline of $\triangle ABC$.
== Proof ==
This is a direct application of the Parallel Transversal Theorem.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.26$
Category:Triangles
Line Perpendicular to Two Intersecting Lines is Perpendicular to their Plane

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be two straight lines which cut one another at the point $E$.
Let a straight line $EF$ be set up at right angles to $AB$ and $CD$.
It is to be demonstrated that $EF$ is also at right angles to the plane through $AB$ and $CD$.
Let $AE$, $EB$, $CE$ and $ED$ be cut off equal to one another.
Let $AD$ and $BC$ be drawn.
Let a straight line $GEH$ be drawn arbitrarily across from $AD$ and $BC$ through $E$ such that $G$ lies on $AD$ and $H$ lies on $BC$.
Let $FA$, $FG$, $FD$, $FC$, $FH$ and $FB$ be joined from the point $F$.
We have that $AE$ and $ED$ are equal to the two straight lines $CE$ and $EB$.
From :
:$\angle AED = \angle CEB$
Therefore the base $AD$ of $\triangle AED$ equals the base $BC$ of $\triangle CEB$.
Therefore by :
:$\angle DAE = \angle EBC$
But from :
:$\angle AEG = \angle BEH$
Therefore $\triangle AGE$ and $\triangle BEH$ are two triangles which have:
:two angles equal to two angles respectively
:one side $AE$ equal to one side $EB$, that adjacent to the equal angles.
Therefore from :
:the remaining sides of $\triangle AGE$ and $\triangle BEH$ will also be equal.
Therefore $GE = EH$ and $AG = BH$
We have that:
: $AE = EB$
while $FE$ is common and at right angles.
Therefore by :
:$FA = FB$
For the same reason:
:$FC = FD$
We have that:
:$AD = CB$
and:
:$FA = FB$
Therefore $FA = FB$ and $AD = BC$
But $FD$ is proved equal to $FC$.
Therefore by :
:$\angle FAD = \angle FBC$
Similarly:
:$AG = BH$
and:
:$FA = FB$
and so $FA = AG$ and $FB = BH$.
We also have that:
:$\angle FAG = \angle FBH$
and so from :
:$FG = FH$
We have that $GE = EH$.
We also have that $EF$ is common.
So the two sides $GE$ and $EF$ are equal to the two sides $HE$ and $EF$.
The side $FG$ equals the side $FH$.
Therefore from :
:$\angle GEF = \angle HEF$
Therefore each of $\angle GEF$ and $\angle HEF$ is right.
Therefore $FE$ is at right angles to $GH$ drawn at random through $E$.
Similarly it can be proved that $FE$ will make right angles with all the straight lines which meet it and are in the plane of reference.
So from :
:$FE$ is at right angles to the plane of reference.
But the plane of reference is the plane through the straight lines $AB$ and $CD$.
Hence the result.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Perpendiculars
Category:Planes
Line at Right Angles to Diameter of Circle

== Theorem ==
=== Porism ===
== Proof ==
:
Let $ABC$ be a circle about $D$ as center and $AB$ as diameter.
Suppose that a line from $A$ at right angles to the diameter falls within the circle, e.g. at $AC$.
Let $DC$ be joined.
Since $DA = DC$ we have that $\angle DAC = \angle ACD$.
But by hypothesis the angle $DAC$ is a right angle.
Therefore $\angle ACD$ is also a right angle.
So in $\triangle ACD$, the two angles $\angle DAC$ and $\angle ACD$ equal two right angles.
From Two Angles of Triangle are Less than Two Right Angles this is impossible.
Therefore the straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle.
Let this line be as $AE$.
Suppose there were another straight line $AF$ interposed between the circumference of the circle and $AE$.
Let $DG$ be drawn perpendicular to $AF$ from $D$.
Since $\angle AGD$ is a right angle, the angle $\angle DAG$ is less than a right angle by Two Angles of Triangle are Less than Two Right Angles.
So $AD > DG$ from Greater Angle of Triangle Subtended by Greater Side.
But $AD = DH$ and so $DH > DG$, which is impossible.
Therefore another straight line cannot be interposed into the space between the straight line and the circumference of the circle.
Suppose there were a rectilineal angle greater than:
:that contained by the straight line $AB$ and the arc $CHA$ of the circle
and:
:any rectilineal angle less than the angle contained by the arc $CHA$ and the straight line $AE$.
Then into the space between the circumference and the straight line $AE$ another straight line can be interposed such as will make an angle contained by straight lines greater than the angle contained by the straight line $AB$ and the arc $CHA$.
Also, another angle contained by the straight lines $AB$ and $AE$ which is less than the angle contained by the arc $CHA$ and the straight line $AE$.
But such a straight line cannot be interposed.
Therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line $AB$ and the arc $CHA$.
Nor will there be any any acute angle contained by straight lines which is less than the angle contained by the arc $CHA$ and the straight line $AE$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Line from Bisector of Side of Parallelogram to Vertex Trisects Diagonal

== Theorem ==
Let $ABCD$ be a parallelogram.
Let $E$ be the midpoint of $AD$.
Then the point at which the line $BE$ meets $AC$ trisects $AC$.
== Proof ==
:
Let the given intersection be at $F$.
We have that $E$ is the midpoint of $AD$.
Thus:
Since $\vec {AB} + \vec {BC}$ we have $\vec {AF} = n \paren {\vec {AB} + \vec {BC} }$ where $0 \le n \le 1$.
But:
:$\vec {AB} + \vec {BF} = \vec {AF}$
That is:
:$\vec {AB} + m \paren {\dfrac {\vec {BC} } 2 - \vec {AB} } = n \paren {\vec {AB} + \vec {BC} }$
That is:
:$\paren {1 - m - n} \vec {AB} + \paren {\dfrac m 2 - n} \vec {AB} = 0$
These have a simultaneous solution because $\vec {AB}$ and $\vec {AD}$ are neither coincident nor parallel.
So:
:$1 - m - n = 0, \dfrac m 2 - n = 0 \implies m = 2 n$
Hence $3n = 1$ and so:
:$n = \dfrac 1 3$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $69$
Category:Parallelograms
Line from Vertex of Triangle to Incenter is Angle Bisector

== Theorem ==
Let $\triangle ABC$ be a triangle.
Let $D$ be a point in the interior of $\triangle ABC$.
Then:
:$AD$ is the angle bisector of $A$
:$BD$ is the angle bisector of $B$
:$CD$ is the angle bisector of $C$
:
:$D$ is the incenter of $\triangle ABC$.
== Proof ==
:
=== Necessary Condition ===
Let $D$ be the incenter of $\triangle ABC$.
Let $E$, $F$ and $G$ be the points on $AB$, $BC$ and $AC$ tangent to the incircle $\bigcirc EFG$ of $\triangle ABC$.
We have that:
:$ED = DF$, as both equal the radius of $\bigcirc EFG$
:$\angle BED = \angle BFD$, a right angle
:$BD$ is common.
From Pythagoras's Theorem we have that:
Hence by Triangle Side-Side-Side Congruence:
:$\triangle EBD = \triangle FBD$
So:
:$\angle EBD = \angle FBD$
and it is seen that $BD$ is the angle bisector of $B$.
By the same reasoning, :
:$AD$ is the angle bisector of $A$
and
:$CD$ is the angle bisector of $C$.
Hence the result.
=== Sufficient Condition ===
Let:
:$AD$ be the angle bisector of $A$
:$BD$ be the angle bisector of $B$
:$CD$ be the angle bisector of $C$
It follows from Inscribing Circle in Triangle that $D$ is the incenter of $\triangle ABC$.
Category:Incircles of Triangles
Category:Angle Bisectors
Line in Plane is Straight iff Slope is Constant

== Theorem ==
Let $\LL$ be a curve which can be embedded in the plane.
Then $\LL$ is a straight line it is of constant slope.
== Proof ==
Let $L$ be embedded in the cartesian plane.
The slope of $\LL$ at a point $p = \tuple {x, y}$ is defined as being its derivative at $p$ $x$:
:$\grad p = \dfrac {\d y} {\d x}$
:
Let $\LL$ be a straight line.
Let $\triangle ABC$ and $\triangle DEF$ be right triangles constructed so that:
:$A, B, D, E$ are on $\LL$
:$AC$ and $DF$ are parallel to the $x$-axis
:$BC$ and $EF$ are parallel to the $y$-axis.
From Parallelism implies Equal Corresponding Angles:
:$\angle ABC = \angle DEF$
and:
:$\angle BAC = \angle EDF$
Also we have that $\angle ACB = \angle DFE$ and are right angles.
Thus $\triangle ABC$ and $\triangle DEF$ are similar.
Thus:
:$\dfrac {BC} {AC} = \dfrac {EF} {DF}$
That is, the slope of $\LL$ between $A$ and $B$ is the same as the slope of $\LL$ between $D$ and $E$.
The argument reverses.
== Sources ==
* 
Category:Straight Lines
Line joining Points on Parallel Lines is in Same Plane

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be two straight lines which are parallel.
Consider the plane in which they lie to be the plane of reference.
Let $E$ and $F$ be taken at random on both of them.
Then the straight line $EF$ is to be demonstrated to lie in the same plane as $AB$ and $CD$.
Suppose to the contrary that $EF$ is in a more elevated plane: $EGF$.
Let a plane be drawn through $EGF$.
From , the common section of this plane with the plane of reference is a straight line passing through $E$ and $F$.
Therefore $EF$ and $EGF$ enclose an area, which is impossible.
Therefore the straight line $EF$ joining $E$ and $F$ is in the same plane as $AB$ and $CD$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Category:Parallel Lines
Line of Electric Force is Continuous

== Theorem ==
Let $\mathbf E$ be an electric field.
Let $L$ be a line of force within $\mathbf E$.
Then, except for the positive charge or negative charge at which $L$ may start or end, $L$ is continuous.
== Proof ==
Follows directly from Vector Line is Continuous.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field
Category:Lines of Electric Force
Line through Two Diagonal Points of Complete Quadrangle is Polar Line of Third Diagonal Point

== Theorem ==
Let $Q$ be a complete quadrangle inscribed in a conic section $\CC$.
Let $R$, $S$ and $T$ be the diagonal points of $Q$.
Let $\LL$ be the straight line through $R$ and $S$.
Then $\LL$ is the polar of $T$ $\CC$.
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
Category:Line through Two Diagonal Points of Complete Quadrangle is Polar Line of Third Diagonal Point
Category:Diagonal Points
Category:Complete Quadrangles
Category:Conic Sections
Linear Bound Lemma

== Theorem ==
Let $G_n$ be a simple connected planar graph with $n$ vertices.
Then:
:$m \le 3 n − 6$
where $m$ is the number of edges.
== Proof ==
Let $f$ denote the number of faces of $G_n$. 
Let $\sequence {s_i}_{i \mathop = 1}^f$ be a sequence of regions of a planar embedding of $G_n$. 
Consider the sequence $\sequence {r_i}_{i \mathop = 1}^f$ where $r_i$ denotes the number of boundary edges of $s_i$. 
Since $G$ is simple, then by the definition of planar embedding: 
:every region has at least $3$ boundary edges
:every edge is a boundary edge of at most two regions in the planar embedding.
Using this two facts, we can find the boundary of $\ds \sum_{i \mathop = 1}^f r_i$ as:
:$3 f \le \ds \sum_{i \mathop = 1}^f r_i \le 2m$
Now, as $f \le \dfrac 2 3 m$:
Category:Graph Theory
Category:Named Theorems
Linear Bound between Complex Function and Derivative

== Theorem ==
Let $f: D \to \C$ be a complex-differentiable function, where $D \subseteq \C$ is an open set.
Let $z_0 \in D$.
Let $\epsilon \in \R_{>0}$.
Then there exists $r \in \R_{>0}$ such that for all $z \in \map {B_r} {z_0}$:
:$\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
where $\map {B_r} {z_0}$ denotes the open ball with center $z_0$ and radius $r$.
== Proof ==
Given $\epsilon > 0$, we can find $r > 0$ by definition of limit such that for all $z \in \map {B_r} {z_0}$:
:$\dfrac {\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } } {\size {z - z_0} } < \epsilon$
We rearrange to:
:$\size {\map f z - \map f {z_0} - \map {f'} {z_0} \paren {z - z_0} } < \epsilon \size {z - z_0}$
== Sources ==
* $\S 3.1$
Category:Complex Differential Calculus
Linear Code/Examples

== Examples of Linear Codes ==
=== Elements of $\map V {3, 2}$ ===
=== Elements of $\map V {2, 3}$ ===
=== Linear $\tuple {2, 2}$-code of $\map V {3, 2}$ ===
=== Linear $\tuple {1, 1}$-code of $\map V {2, 3}$ ===
== Examples of Non-Linear Codes ==
Category:Examples of Linear Codes
Linear Code/Examples/(1,1) of V(2,3)

== Example of Linear Code ==
The set:
:$C := \set {11, 22, 33}$
is a linear $\tuple {2, 1}$-code of elements of $\Z_3$.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.4$
Category:Examples of Linear Codes
Linear Code/Examples/(2,2) of V(3,2)

== Example of Linear Code ==
The set:
:$C := \set {000, 001, 010, 011}$
is a linear $\tuple {2, 2}$-code of elements of $\Z_2$.
== Proof ==
:$(C \, 1): \quad$ Subtracting any of the codewords from any other gives a codeword beginning with $0$.
:$(C \, 2): \quad$ As $0$ and $1$ are the only elements of $\Z_2$, then any multiple of any code is another element of $C$.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.3$
Category:Examples of Linear Codes
Linear Code/Examples/Non-Linear Code/Subset of V(6,2)

== Example of Non-Linear Code ==
Let $\map V {6, 2}$ denote the set of sequences of length $6$ of elements of $\Z_2$.
For all $x \in \Z_p$, let $\overline x$ denote $1 + x \pmod 2$ so that:
:$\overline 0 = 1$
:$\overline 1 = 0$
Let $C \subseteq \map V {6, 2}$ be the set of elements of $\map V {6, 2}$ of the form:
:$x y z \overline x \overline y \overline z$
Then $C$ is not a linear code.
== Proof ==
The elements of $C$ are:
:$000111, 001110, 010101, 011100, 100011, 101010, 110001, 111000$
The minimum distance of $C$ is $2$.
We immediately note that:
:$000111 + 001110 = 001001$
but that $001001 \notin C$.
So $C$ is not a linear code by definition.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $1$
Category:Examples of Linear Codes
Linear Code/Examples/V(2,3)

== Example of Linear Code ==
The master code $\map V {2, 3}$ of length $2$ of elements of $\Z_3$ consists of:
:$\map V {2, 3} = \set {00, 01, 02, 10, 11, 12, 20, 21, 22}$
== Sources ==
* : Chapter $6$: Error-correcting codes
Category:Examples of Linear Codes
Linear Code/Examples/V(3,2)

== Example of Linear Code ==
The master code $\map V {3, 2}$ of length $3$ of elements of $\Z_2$ consists of:
:$\map V {3, 2} = \set {000, 001, 010, 011, 100, 101, 110, 111}$
== Sources ==
* : Chapter $6$: Error-correcting codes
Category:Examples of Linear Codes
Linear Combination Operator on Real Numbers is Entropic

== Theorem ==
Let $a, b \in \R$ be real numbers.
Let $\odot$ be the operation on $\R$ defined as:
:$\forall x, y \in \R: x \odot y := a x + b y$
Then $\odot$ is an entropic operation.
== Proof ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.12 \ \text{(d)}$
Category:Linear Combinations
Category:Real Numbers
Category:Examples of Entropic Operations
Linear Combination of Absolutely Continuous Function is Absolutely Continuous

== Theorem ==
Let $\alpha, \beta$ be real numbers.
Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $\alpha f + \beta g$ is absolutely continuous.
== Proof ==
From Multiple of Absolutely Continuous Function is Absolutely Continuous, we have: 
:$\alpha f$ and $\beta g$ are absolutely continuous.
We therefore have, by Sum of Absolutely Continuous Functions is Absolutely Continuous:
:$\alpha f + \beta g$ is absolutely continuous.
Category:Absolutely Continuous Real Functions
Linear Combination of Balanced Sets is Balanced

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $\family {E_\alpha}_{\alpha \mathop \in A}$ be an $A$-indexed family of balanced sets. 
Let $\lambda_\alpha \in \GF$ for each $\alpha \mathop \in A$. 
Then:
:$\ds \sum_{\alpha \mathop \in A} \lambda_\alpha E_\alpha$ is balanced.
== Proof ==
Let $s \in \C$ have $\cmod s \le 1$.
Then, we have:
So:
:$\ds \sum_{\alpha \mathop \in A} \lambda_\alpha E_\alpha$ is balanced.
Category:Balanced Sets
Linear Combination of Bounded Mappings on Normed Vector Space is Bounded

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,}_R}$ be a normed division ring.
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $R$. 
Let $S$ be a set.
Let $f, g : S \to X$ be bounded mappings.
Let $\lambda \in R$. 
Then $f + \lambda g$ is bounded.
== Proof ==
Since $f$ is bounded, there exists a real number $M_1 > 0$ such that:
:$\norm {\map f x} \le M_1$ for each $x \in X$. 
Since $g$ is bounded, there exists a real number $M_2 > 0$ such that:
:$\norm {\map g x} \le M_2$ for each $x \in X$. 
Then from , we have:
:$\norm {\map f x + \lambda \map g x} \le \norm {\map f x} + \norm \lambda_R \norm {\map g x}$
for each $x \in X$.
So:
:$\norm {\map f x + \lambda \map g x} \le M_1 + \norm \lambda_R M_2$ for each $x \in X$.
Since $M_1 + \norm \lambda_R M_2 > 0$, we have that $f + \lambda g$ is bounded.
Category:Bounded Mappings on Normed Vector Spaces
Linear Combination of Cauchy Sequences in Topological Vector Space is Cauchy Sequence

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be Cauchy sequences in $X$.
Let $\lambda, \mu \in \GF$.
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is a Cauchy sequence in $X$.
== Proof ==
From Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n}_{n \in \N}$ and $\sequence {\mu y_n}_{n \in \N}$ are Cauchy in $X$.
From Sum of Cauchy Sequences in Topological Vector Space is Cauchy Sequence, $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ is Cauchy.
Category:Cauchy Sequences in Topological Vector Spaces
Linear Combination of Complex-Valued Functions Vanishing at Infinity Vanishes at Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f, g : X \to \C$ be complex-valued functions vanishing at infinity.
Let $\lambda \in \C$.
Then $f + \lambda g$ vanishes at infinity.
== Proof ==
If $\lambda = 0$, we are done immediately. 
Take $\lambda \ne 0$. 
Let $\epsilon > 0$. 
Since $f$ vanishes at infinity, there exists a compact set $F_1 \subseteq X$ such that:
:$\cmod {\map f x} < \dfrac \epsilon 2$ for each $x \in X \setminus F_1$.
Since $g$ vanishes at infinity, there exists a compact set $F_2 \subseteq X$ such that:
:$\cmod {\map g x} < \dfrac \epsilon {2 \cmod \lambda}$ for each $x \in X \setminus F_2$. 
From Union of Two Compact Sets is Compact, $F_1 \cup F_2$ is compact and:
:$\cmod {\map f x} < \dfrac \epsilon 2$ and $\cmod {\map g x} < \dfrac \epsilon {2 \cmod \lambda}$ for each $x \in X \setminus \paren {F_1 \cup F_2}$.
Then, we have for $x \in X \setminus \paren {F_1 \cup F_2}$:
Since again $F_1 \cup F_2$ is compact, we have that $f + \lambda g$ vanishes at infinity.
Category:Complex-Valued Functions Vanishing at Infinity
Linear Combination of Complex-Valued Functions Vanishing in Neighborhood of Infinity Vanishes in Neighborhood of Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f, g : X \to \C$ be complex-valued functions vanishing in a neighborhood of infinity.
Let $\lambda \in \C$.
Then $f + \lambda g$ vanishes in a neighborhood of infinity.
== Proof ==
Since $f$ vanishes in a neighborhood of infinity, there exists a compact set $F_1 \subseteq X$ such that:
:$\map f x = 0$ for each $x \in X \setminus F_1$.
Since $g$ vanishes in a neighborhood of infinity, there exists a compact set $F_2 \subseteq X$ such that:
:$\map g x = 0$ for each $x \in X \setminus F_2$.
From Union of Two Compact Sets is Compact:
:$F_1 \cup F_2$ is compact.
Further:
:$\map f x = 0$ and $\map g x = 0$ for each $x \in X \setminus \paren {F_1 \cup F_2}$. 
So we have:
:$\map f x + \lambda \map g x = 0$ for each $x \in X \setminus \paren {F_1 \cup F_2}$. 
Recalling that $F_1 \cup F_2$ is compact, we have that $f + \lambda g$ vanishes in a neighborhood of infinity.
Category:Complex-Valued Functions Vanishing in Neighborhood of Infinity
Linear Combination of Complex Integrals

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f, g: \closedint a b \to \C$ be complex Riemann integrable functions over $\closedint a b$.
Let $\lambda, \mu \in \C$ be complex constants.
Then:
:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
== Proof ==
First, we prove the result for addition only without multiplying by $\lambda, \mu$:
 
Next, we show the result for only one complex integral multiplied by a constant $\lambda$.
By definition of complex number, we can find $\lambda_x, \lambda_y \in \R$ so $\lambda = \lambda_x + i \lambda_y$.
Then:
It follows from the results above that:
== Sources ==
* $\S 2.1$
Category:Complex Analysis
Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous

== Theorem ==
Let $X$ be a topological space. 
Let $K$ be a topological field.
Let $Y$ be a topological vector space over $K$. 
Let $n \in \N$.
Let $f_1, \ldots, f_n : X \to Y$ be continuous functions. 
Let $\alpha_1, \ldots, \alpha_n \in K$. 
Then: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k f_k$ is a continuous function.
== Proof ==
We do induction on the number of functions $n$:
=== Basis for the Induction ===
For each $\lambda \in K$, define $c_\lambda : Y \to Y$ by: 
:$\map {c_\lambda} y = \lambda y$
for each $y \in Y$.
From Dilation Mapping on Topological Vector Space is Continuous, $c_\lambda$ is continuous for each $\lambda \in K$.
Then we have: 
:$\alpha_1 f_1 = c_{\alpha_1} \circ f_1$
Since $f_1$ is continuous, $\alpha_1 f_1$ is continuous by Composite of Continuous Mappings is Continuous.
So shown for basis for the induction.
=== Induction Hypothesis ===
This is the induction hypothesis:
:$\ds \sum_{k \mathop = 1}^n \alpha_k f_k$ is a continuous function.
Now we want to show that if $f_{n + 1} : X \to Y$ is another continuous function, we have: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \alpha_k f_k$ is continuous.
=== Induction Step ===
This is the induction step.
Write: 
:$\ds s_n = \sum_{k \mathop = 1}^n \alpha_k f_k$
From the base case: 
:$\alpha_{n + 1} f_{n + 1}$ is continuous.
Define $f : X \to Y \times Y$ by: 
:$\map f x = \tuple {\map {s_n} x, \alpha_{n + 1} f_{n + 1} }$ 
for each $x \in X$.
From Continuous Mapping to Product Space, $f$ is continuous.
Define $g : Y \times Y \to Y$ by: 
:$\map g {x, y} = x + y$
for each $x, y \in Y$.
Since $Y$ is a topological vector space, $g$ is continuous.
From Continuous Mapping to Product Space, $g \circ f$ is continuous.
We have: 
:$\ds g \circ f = \sum_{k \mathop = 1}^n \alpha_k f_k + \alpha_{n + 1} f_{n + 1} = \sum_{k \mathop = 1}^{n + 1} \alpha_k f_k$
So: 
:$\ds \sum_{k \mathop = 1}^{n + 1} \alpha_k f_k$ is continuous.
The result follows by induction.
</onlyinclude>
Category:Topological Vector Spaces
Linear Combination of Contour Integrals

== Theorem ==
Let $C$ be a contour in $\C$.
Let $f, g: \Img C \to \C$ be continuous complex functions, where $\Img C$ denotes the image of $C$.
Let $\lambda, \mu \in \C$ be complex constants.
Then:
:$\ds \int_C \paren {\lambda \map f z + \mu \map g z} \rd z = \lambda \int_C \map f z \rd z + \mu \int_C \map g z \rd z$
== Proof ==
By definition of contour, $C$ is a finite sequence $C_1, \ldots, C_n$ of directed smooth curves.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Then:
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals