nyuuzyou/wikis huggingface dataset:
Conversion per Accidens

== Theorem ==
Consider the categorical statements:
Then:
:$\map {\mathbf A} {S, P} \implies \map {\mathbf I} {P, S}$
:
:$\exists x: \map S x$
Using the symbology of predicate logic:
:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map P x \land \map S x} }$
This law has the traditional name conversion per accidens of $\mathbf A$.
Thus the $\mathbf A$ form converts per accidens to the $\mathbf I$ form.
== Proof ==
From Universal Affirmative implies Particular Affirmative iff First Predicate is not Vacuous:
:$\exists x: \map S x \iff \paren {\paren {\forall x: \map S x \implies \map P x} \implies \paren {\exists x: \map S x \land \map P x} }$
From Law of Simple Conversion of I:
:$\paren {\exists x: \map S x \land \map P x} \implies \paren {\exists x: \map P x \land \map S x}$
Hence the result.
== Also defined as ==
Some sources gloss over the possibility of $\map S x$ being vacuous and merely report that the universal affirmative $\map {\mathbf A} {S, P}$ implies the particular affirmative $\map {\mathbf I} {P, S}$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism
Category:Categorical Statements
Converting Decimal Expansion of Rational Number to Fraction

== Theorem ==
Let $x \in \Q$ be a rational number.
Let the decimal expansion of $x$ be:
:$0 \cdotp a_1 a_2 \ldots a_m \dot b_1 b_2 \ldots \dot b_n$
where $a_i: i \in \set {1, 2, \ldots, m}$ and $b_j: j \in \set {1, 2, \ldots, n}$ be the digits in the base $10$ expansion of $x$.
Then $x$ can be expressed as the following fraction:
:$x = \dfrac {a_1 a_2 \ldots a_m b_1 b_2 \ldots b_n - a_1 a_2 \ldots a_m} {10^m \times \paren {10^n - 1} }$
== Proof ==
First we note that:
== Examples ==
Category:Converting Decimal Expansion of Rational Number to Fraction
Category:Recurring Decimals
Category:Rational Numbers
Category:Decimal Notation
Category:Proof Techniques
Converting Decimal Expansion of Rational Number to Fraction/Examples

== Examples of Converting Decimal Expansion of Rational Number to Fraction ==
=== Decimal Number $0 \cdotp 7 \dot 1 82 \dot 8$ ===
=== Decimal Number $0 \cdotp 14 \dot 1 \dot 6$ ===
Category:Examples of Converting Decimal Expansion of Rational Number to Fraction
Convex Absorbing Set contained between Sets in terms of Minkowski Functional

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$.
 
Let $C$ be a convex absorbing set.
Let $\mu_C$ be the Minkowski functional of $C$. 
Then we have: 
:$\set {x \in X : \map {\mu_C} x < 1} \subseteq C \subseteq \set {x \in X : \map {\mu_C} x \le 1}$
== Proof ==
Let $x \in X$ be such that $\map {\mu_C} x < 1$.
Then: 
:$\inf \set {t > 0 : \dfrac x t \in C} < 1$
So there exists $t < 1$ such that $x \in t C$.
From Absorbing Set in Vector Space contains Zero Vector, we have that ${\mathbf 0}_X \in C$. 
So, we have, since $C$ is convex: 
:$x = x + \paren {1 - t} {\mathbf 0}_X \in t C + \paren {1 - t} C \subseteq C$
So $x \in C$.
So we have: 
:$\set {x \in X : \map {\mu_C} x < 1} \subseteq C$
Now let $x \in C$.
We clearly have: 
:$1 \in \set {t > 0 : \dfrac x t \in C}$
So:
:$\map {\mu_C} x = \inf \set {t > 0 : \dfrac x t \in C} \le 1$
Hence we obtain the inclusion: 
:$C \subseteq \set {x \in X : \map {\mu_C} x \le 1}$
Category:Minkowski Functionals
Convex Combination/Examples/Arbitrary Example 1

== Example of Convex Combination ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be vectors of a vector space $V$.
Then:
:$0.1 \mathbf a + 0.3 \mathbf b + 0.6 \mathbf c$ is a convex combination of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
== Sources ==
* 
Category:Examples of Convex Combinations
Convex Cone is Convex Set

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$.
Let $P \subseteq X$ be a convex cone in $X$.
Then $P$ is convex.
== Proof ==
Let $x, y \in P$. 
Let $t \in \closedint 0 1$ so that:
:$t \ge 0$ and $1 - t \ge 0$.
Since $P$ is a cone, we have:
:$t x \in P$ and $\paren {1 - t} y \in P$.
Since $P$ is a convex cone, we have:
:$t x + \paren {1 - t} y \in P$
So $P$ is convex.
== Sources ==
* : $1.1$: Basic notions
Category:Convex Cones
Convex Hull is Smallest Convex Set containing Set

== Theorem ==
Let $X$ be a vector space over $\R$.
Let $U \subseteq X$ be non-empty.
Let $\map {\operatorname {conv} } U$ be the convex hull of $U$.
Then $\map {\operatorname {conv} } U$ is the smallest convex subset of $X$ containing $U$ in the sense that: 
:$\map {\operatorname {conv} } U$ is convex and if $K \subseteq X$ is a convex subset with $U \subseteq K$, we have that $\map {\operatorname {conv} } U \subseteq K$.
=== Corollary ===
== Proof ==
We have: 
:$\ds \map {\operatorname {conv} } U = \set {\sum_{j \mathop = 1}^n \lambda_j u_j : n \in \N, \, u_j \in U \text { and } \lambda_j \in \R_{> 0} \text { for each } j, \, \sum_{j \mathop = 1}^n \lambda_j = 1}$
Considering $n = 1$ and $\lambda_1 = 1$, we obtain: 
:$u \in \map {\operatorname {conv} } U$ for each $u \in U$. 
So:
:$U \subseteq \map {\operatorname {conv} } U$
Now we prove that $\map {\operatorname {conv} } U$ is convex.
Let $u, v \in \map {\operatorname {conv} } U$.
Then there exists $u_1, u_2, \ldots, u_n \in U$, $v_1, v_2, \ldots, v_m$ and $\lambda_1, \lambda_2, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in \R_{> 0}$ such that: 
:$\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$
and:
:$\ds v = \sum_{j \mathop = 1}^n \mu_j v_j$
with:
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
and:
:$\ds \sum_{j \mathop = 1}^n \mu_j = 1$ 
Now let $t \in \closedint 0 1$.
We have: 
:$\ds t u + \paren {1 - t} v = \sum_{i \mathop = 1}^n t \lambda_i u_i + \sum_{j \mathop = 1}^m \paren {1 - t} \mu_j \lambda_j$ 
We have: 
with $t \lambda_i \ge 0$ and $\paren {1 - t} \mu_j \ge 0$ for each $i, j$.
Since we have $u_i \in U$ for each $i$ and $v_j \in U$ for each $j$, we have: 
:$t u + \paren {1 - t} v \in \map {\mathrm {conv} } U$
So $\map {\mathrm {conv} } U$ is convex.
=== Lemma ===
Hence the result.
== Sources ==
* $21.5$: The Convex Hull
Category:Convex Hull is Smallest Convex Set containing Set
Category:Convex Hulls
Category:Convex Sets (Vector Spaces)
Convex Hull is Smallest Convex Set containing Set/Corollary

== Theorem ==
Let $X$ be a vector space over $\R$.
Let $K \subseteq X$ be non-empty.
Then: 
:$K$ is convex $\map {\operatorname {conv} } K = K$
where $\map {\operatorname {conv} } K$ denotes the convex hull of $K$.
== Proof ==
=== Sufficient Condition ===
Suppose that: 
:$\map {\operatorname {conv} } K = K$
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$\map {\operatorname {conv} } K$ is convex.
So:
:$K$ is convex.
=== Necessary Condition ===
Suppose that: 
:$K$ is convex.
From Convex Hull is Smallest Convex Set containing Set, we have: 
:$K \subseteq \map {\operatorname {conv} } K$
Note that $K$ is a convex set with $K \subseteq K$, from Set is Subset of Itself.
So Convex Hull is Smallest Convex Set containing Set also gives: 
:$\map {\operatorname {conv} } K \subseteq K$
so:
:$\map {\operatorname {conv} } K = K$
Category:Convex Hull is Smallest Convex Set containing Set
Convex Hull is Smallest Convex Set containing Set/Lemma

== Lemma ==
Let $X$ be a vector space over $\R$.
Let $U \subseteq X$ be non-empty.
Let $\map {\operatorname {conv} } U$ be the convex hull of $U$.
Let $K \subseteq X$ be a convex set with $U \subseteq K$.
Then: 
:$\map {\operatorname {conv} } U \subseteq K$
== Proof ==
From the definition of the convex hull, we have: 
:$\ds \map {\operatorname {conv} } U = \set {\sum_{j \mathop = 1}^n \lambda_j u_j : n \in \N, \, u_j \in U \text { and } \lambda_j \in \R_{> 0} \text { for each } j, \, \sum_{j \mathop = 1}^n \lambda_j = 1}$
For each $n \in \N$, let $\map P n$ be the proposition:
:for each $\lambda_1, \lambda_2, \ldots, \lambda_n \in \R_{> 0}$ with:
::$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
:and $u_1, u_2, \ldots, u_n \in U$, we have: 
::$\ds \sum_{i \mathop = 1}^n \lambda_i u_i \in K$
We will then have that, for each $n \in \N$, $u_1, u_2, \ldots, u_n \in U$ and $\lambda_1, \lambda_2, \ldots, \lambda_n \in \R_{> 0}$ with: 
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
we have: 
:$\ds \sum_{i \mathop = 1}^n \lambda_i u_i \in K$
That is: 
:$\map {\operatorname {conv} } U \subseteq K$
which was the requirement. 
We proceed by induction.
=== Basis for the Induction ===
The case $n = 1$ is precisely the statement that: 
:$\lambda_1 u_1 \in K$
for all $u_1 \in U$ whenever $\lambda_1 = 1$. 
That is: 
:$u_1 \in K$
whenever $u_1 \in U$.
This is precisely the statement that $U \subseteq K$, which is true by hypothesis. 
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:for each $\lambda_1, \lambda_2, \ldots, \lambda_k \in \R_{> 0}$ with:
::$\ds \sum_{i \mathop = 1}^k \lambda_i = 1$
:and $u_1, u_2, \ldots, u_k \in U$, we have: 
::$\ds \sum_{i \mathop = 1}^k \lambda_i u_i \in K$
from which it is to be shown that:
:for each $\lambda_1, \lambda_2, \ldots, \lambda_{k + 1} \in \R_{> 0}$ with:
::$\ds \sum_{i \mathop = 1}^{k + 1} \lambda_i = 1$
:and $u_1, u_2, \ldots, u_{k + 1} \in U$, we have: 
::$\ds \sum_{i \mathop = 1}^{k + 1} \lambda_i u_i \in K$
=== Induction Step ===
Take $\lambda_1, \lambda_2, \ldots, \lambda_{k + 1} \in \R_{> 0}$ with: 
:$\ds \sum_{i \mathop = 1}^{k + 1} \lambda_i = 1$ 
and $u_1, u_2, \ldots, u_{k + 1} \in U$.
We appeal to the convexity of $K$.
Set: 
:$t = 1 - \lambda_{k + 1}$ 
Since we have: 
:$0 < \lambda_i < 1$
for each $i$, we have $0 < t < 1$. 
For each $i \in \set {1, 2, \ldots, k}$ set: 
:$\ds \mu_i = \frac 1 t \lambda_i$
We then have: 
and:
From the induction hypothesis, we have: 
:$\ds \sum_{i \mathop = 1}^k \mu_i u_i \in K$
Since $U \subseteq K$ from hypothesis, we have: 
:$\ds u_{k + 1} \in K$
So, since $K$ is convex, we have: 
:$\ds t \sum_{i \mathop = 1}^k \mu_i u_i + \paren {1 - t} u_{k + 1} \in K$
so:
:$\ds \sum_{i \mathop = 1}^{k + 1} \lambda_i u_i \in K$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Convex Hull is Smallest Convex Set containing Set
Category:Proofs by Induction
Convex Hull preserves Subsets

== Theorem ==
Let $X$ be a vector space over $\R$.
Let $U, V \subseteq X$ be non-empty with $U \subseteq V$. 
Then: 
:$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$
where $\operatorname {conv}$ denotes the convex hull.
== Proof ==
Let: 
:$u \in \map {\operatorname {conv} } U$
From the definition of the convex hull, there exists $u_1, u_2, \ldots, u_n \in U$ and $\lambda_1, \lambda_2, \ldots, \lambda_n \in \R_{> 0}$ with: 
:$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$
such that: 
:$\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$
Since $U \subseteq V$, we have: 
:$u_i \in V$ for each $i$.
So, we have: 
:$u \in \map {\operatorname {conv} } V$ 
So:
:if $u \in \map {\operatorname {conv} } U$ then $u \in \map {\operatorname {conv} } V$.
So, from the definition of subset, we have: 
:$\map {\operatorname {conv} } U \subseteq \map {\operatorname {conv} } V$
Category:Convex Hulls
Convex Real Function Composed with Increasing Convex Real Function is Convex

== Theorem ==
Let $I$ be real interval.
Let $f : I \to \R$ be a convex function.
Let $J$ be a real interval containing the image of $f$.
Let $g : J \to \R$ be a increasing convex function.
Then $g \circ f : I \to \R$ is a convex function.
== Proof ==
Let $x, y \in I$ and $t \in \closedint 0 1$.
Since $f$ is convex, we have: 
:$\map f {t x + \paren {1 - t} y} \le t \map f x + \paren {1 - t} \map f y$ 
Since $g$ is increasing, we then have: 
:$\map {\paren {g \circ f} } {t x + \paren {1 - t} y} \le \map g {t \map f x + \paren {1 - t} \map f y}$
Since $g$ is convex, we then have: 
:$\map g {t \map f x + \paren {1 - t} \map f y} \le t \map g {\map f x} + \paren {1 - t} \map g {\map f y}$ 
so that: 
:$\map {\paren {g \circ f} } {t x + \paren {1 - t} y} \le t \map {\paren {g \circ f} } x + \paren {1 - t} \map {\paren {g \circ f} } y$
That is: 
:$g \circ f : I \to \R$ is a convex function.
Category:Convex Real Functions
Convex Real Function is Continuous

== Theorem ==
Let $f$ be a real function which is convex on the open interval $\openint a b$.
Then $f$ is continuous on $\openint a b$.
== Proof ==
From Convex Real Function is Left-Hand and Right-Hand Differentiable, $f$ is left-hand and right-hand differentiable on $\openint a b$.
From Left-Hand and Right-Hand Differentiable Function is Continuous, $f$ is continuous on $\openint a b$.
== Also see ==
* Concave Real Function is Continuous
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 22$
* : $\S 12.16$
Category:Convex Real Functions
Category:Continuous Real Functions
Convex Real Function is Left-Hand and Right-Hand Differentiable

== Theorem ==
Let $f$ be a real function which is convex on the open interval $\openint a b$.
Then the left-hand derivative $\map {f'_-} x$ and right-hand derivative $\map {f'_+} x$ both exist for all $x \in \openint a b$.
== Proof ==
Let $f$ be convex on $\openint a b$.
Then by definition of convexity:
:$\forall x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
Let $0 < h_1 < h_2$.
Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then:
:$\dfrac {\map f {x + h_1} - \map f x} {h_1} \le \dfrac {\map f {x + h_2} - \map f x} {h_2}$
Hence the function $\map F h = \dfrac {\map f {x + h} - \map f x} h$ increases in some $\openint 0 \delta$.
Thus from Limit of Increasing Function it follows that both $\ds \lim_{h \mathop \to 0^+} \map F h = \map {f'_+} x$ and $\ds \lim_{h \mathop \to 0^-} \map F h = \map {f'_-} x$ exist.
== Also see ==
* Concave Real Function is Left-Hand and Right-Hand Differentiable
== Sources ==
* : $\S 12.14$
Category:Convex Real Functions
Convex Real Function is Measurable/Proof 2

== Theorem ==
== Proof ==
From Convex Real Function is Pointwise Supremum of Affine Functions: Corollary, there exists a countable set $\SS \subseteq \R^2$ such that: 
:$\ds \map f x = \sup_{\tuple {a, b} \mathop \in \SS} \paren {a x + b}$
for each $x \in \R$.
From Linear Function is Continuous, the map $x \mapsto a x + b$ is continuous for each $\tuple {a, b} \in \SS$. 
From Continuous Mapping is Measurable, the map $x \mapsto a x + b$ is measurable.
Since $\SS$ is countable, we can apply Pointwise Supremum of Measurable Functions is Measurable to obtain that $f$ is measurable.
Category:Convex Real Function is Measurable
Convex Real Function is Pointwise Supremum of Affine Functions

== Theorem ==
Let $f : \R \to \R$ be a convex real function. 
Then there exists a set $\SS \subseteq \R^2$ such that: 
:$\ds \map f x = \sup_{\tuple {a, b} \in \SS} \paren {a x + b}$
for each $x \in \R$. 
=== Corollary ===
== Proof ==
Let $x \in \R$. 
We construct an affine function $\phi_x : \R \to \R$ such that: 
:$\map f y \ge \map {\phi_x} y$
and $\map {\phi_x} x = \map f x$.
We will then argue that: 
:$\map f x = \sup \set {\map {\phi_y} x : y \in \R}$
Define a function $g_x : \R \setminus \set x \to \R$ by: 
:$\map {g_x} y = \dfrac {\map f y - \map f x} {y - x}$
for each $y \in \R \setminus \set x$.
From definition 3 of a convex real function, we have that $g_x$ is increasing.
Then from Limit of Increasing Function, we have: 
:$\sup \set {\map {g_x} y : y < x} = c_x < \infty$
and from : 
:$\map {g_x} y \ge c_x$
for $y > x$.
That is, for $y - x > 0$, we have: 
:$\dfrac {\map f y - \map f x} {y - x} \ge c_x$
and for $y - x < 0$, we have: 
:$\dfrac {\map f y - \map f x} {y - x} \le c_x$
So for all $x, y \in \R$ with $x \ne y$ we have:
:$\map f y - \map f x \ge c_x \paren {y - x}$
giving:
:$\map f y \ge \map f x + c_x \paren {y - x}$
Clearly this inequality also holds when $x = y$. 
Now set:
:$\map {\phi_x} y = \map f x + c_x \paren {y - x}$ 
for each $y \in \R$. 
Then we clearly have: 
:$\map {\phi_x} x = \map f x$
while: 
:$\map f x \ge \map {\phi_y} x$ 
for all $x, y \in \R$.
The latter gives that: 
:$\sup \set {\map {\phi_y} x : y \in \R} \le \map f x$
while the former assures that: 
:$\map f x \le \sup \set {\map {\phi_y} x : y \in \R}$ 
So we have: 
:$\map f x = \sup \set {\map {\phi_y} x : y \in \R}$
In particular, setting: 
:$\SS = \set {\tuple {c_y, \map f y - c_y y} : y \in \R}$
we obtain: 
:$\ds \map f x = \sup_{\tuple {a, b} \in \SS} \paren {a x + b}$
for each $x \in \R$.
Category:Convex Real Functions
Category:Convex Real Function is Pointwise Supremum of Affine Functions
Convex Real Function is Pointwise Supremum of Affine Functions/Corollary

== Corollary to Convex Real Function is Pointwise Supremum of Affine Functions ==
Let $f : \R \to \R$ be a convex real function. 
Then tThere exists a countable set $\mathcal S \subseteq \R^2$ such that: 
:$\ds \map f x = \sup_{\tuple {a, b} \in \mathcal S} \paren {a x + b}$
for each $x \in \R$. 
== Proof ==
For each $x \in \R$, construct $c_x$, $\phi_x$ and $g_x$ as in Convex Real Function is Pointwise Supremum of Affine Functions.
That is, define $g_x : \R \setminus \set x \to \R$ to have: 
:$\ds \map {g_x} y = \frac {\map f y - \map f x} {y - x}$
for each $y \in \R \setminus \set x$.
Define also: 
:$c_x = \sup \set {\map {g_x} y : y < x}$
and:
:$\map {\phi_x} y = \map f x + c_x \paren {y - x}$
for each $y \in \R$. 
Then we have: 
:$\map f x = \sup \set {\map {\phi_y} x : y \in \R}$
We will now show:
:$\map f x = \sup \set {\map {\phi_y} x : y \in \Q}$
for each $x \in \R$. 
Note that from Supremum of Subset, we have: 
:$\sup \set {\map {\phi_y} x : y \in \Q} \le \sup \set {\map {\phi_y} x : y \in \R} = \map f x$
From Monotone Convergence Theorem (Real Analysis): Increasing Sequence, it is sufficient to construct a sequence of rational numbers $\sequence {q_n}_{n \in \N}$ such that: 
:$\map {\phi_{q_n} } x \to \map f x$
From Rational Sequence Increasing to Real Number, there exists an increasing sequence of rational numbers with $q_n \to x$.
Since $q_n \le q_{n + 1}$, by Supremum of Subset we have: 
:$\sup \set {\map {g_x} y : y < q_n} \le \sup \set {\map {g_x} y : y < q_{n + 1} }$
and so:
:$c_{q_n} \le c_{q_{n + 1} }$
for each $n \in \N$.
We also have, since $q_n \le x$: 
:$\sup \set {\map {g_x} y : y < q_n} \le \sup \set {\map {g_x} y : y < x}$ 
and so: 
:$c_{q_n} \le c_x$
for each $n \in \N$. 
If $c_{q_1} \ge 0$, then we have:
:$c_{q_n} \ge 0$ for each $n \in \N$
and hence: 
:$\size {c_{q_n} } \le \size {c_x}$ for each $n \in \N$. 
If $c_{q_1} < 0$, then $\sequence {\size {c_{q_n} } }_{n \in \N}$ is a decreasing sequence until we hit an $m \in \N$ such that $c_{q_m} \ge 0$. 
If no such $m \in \N$ exists, then the sequence is simply decreasing. 
We conclude from these observations that: 
:$\size {c_{q_n} } \le \max \set {\size {c_{q_1} }, \size {c_x} }$ for each $n \in \N$. 
So, we now have:
We then have: 
:$\size {x - q_n} \to 0$ as $n \to \infty$
and so: 
:$\map {\phi_{q_n} } x \to \map f x$
by the Squeeze Theorem.
We therefore have: 
:$\map f x = \sup \set {\map {\phi_y} x : y \in \Q}$
and so setting:
:$\mathcal S = \set {\tuple {c_y, \map f y - c_y y} : y \in \Q}$
we obtain: 
:$\ds \map f x = \sup_{\tuple {a, b} \in \mathcal S} \paren {a x + b}$
for each $x \in \R$, for a countable $\mathcal S$. 
Category:Convex Real Function is Pointwise Supremum of Affine Functions
Convex Set Characterization (Order Theory)

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be an ordered set.
Let $C \subseteq S$.
== Proof ==
=== $(2)$ implies $(1)$ ===
Follows from Upper Section is Convex, Lower Section is Convex, and Intersection of Convex Sets is Convex Set (Order Theory).
=== $(1)$ implies $(3)$ ===
Let $C$ be a convex set in $S$.
Let $U$ and $L$ be the upper and lower closures of $C$, respectively.
Since $C \subseteq U$ and $C \subseteq L$:
:$C \subseteq U \cap L$.
Let $p \in U \cap L$.
Then $a \preceq p \preceq b$ for some $a, b \in C$.
Since $C$ is convex, $p \in C$.
Since this holds for all $p \in U \cap L$:
:$U \cap L \subseteq C$.
Since we know that $C \subseteq U \cap L$, $C = U \cap L$.
=== $(3)$ implies $(2)$ ===
Follows from Upper Closure is Upper Section and Lower Closure is Lower Section.
Category:Convex Sets (Order Theory)
Convex Set is Contractible

== Theorem ==
Let $V$ be a topological vector space over $\R$ or $\C$.
Let $A\subset V$ be a convex subset.
Then $A$ is contractible.
== Proof ==
Let $x_0 \in A$.
Define $H : A \times \closedint 0 1 \to A$ by:
:$\map H {x, t} = t x_0 + \paren {1 - t} x$
Note that by the definition of a convex set, $\Cdm H$ is also a convex set.
Let $I_A$ be defined as the identity mapping: 
:$I_A: A \to A$
Such that:
:$I_A = \set {\tuple {x, x}: x \in A}$
Let $S$ and $T$ be sets. Let $x_0$ be defined as the constant map:
:$x_0: S \to T$ defined as:
:$0 \in T: x_0: S \to T: \forall x \in S: \map {x_0} x = 0$
The map $H$ yields a (free) homotopy between the identity map:
:$I_A$
and the constant map:
:$x_0$
By the assumption of convexity for $A$, $H$ takes values in $A$.
Since $H$ is a polynomial separately in $x, t$, it is a continuous function.
And:
:$\map H {-, 0} = I_A$
:$\map H {-, 1} \equiv x_0$ (the constant function) 
Thus, $H: I_A \simeq c_{x_0}$.
It follows that, the identity map $\operatorname{id}_A$ is homotopic to the constant map $A \to A$.
Hence, by definition of a contractible space, this implies that:
:$A$ is contractible
as required.
Category:Vector Spaces
Category:Topology
Category:Convex Sets (Vector Spaces)
Convex Set is Path-Connected

== Theorem ==
Let $V$ be a topological vector space over $\R$ or $\C$.
A convex subset is path-connected.
== Proof ==
The result follows from:
* Convex Set is Star Convex Set
* Star Convex Set is Path-Connected
Category:Topological Vector Spaces
Category:Convex Sets (Vector Spaces)
Category:Path-Connected Spaces
Convex Set is Simply Connected

== Theorem ==
Let $\struct {V, \tau}$ be a topological vector space over $\R$ or $\C$.
Let $A \subseteq V$ be a non-empty convex set.
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Then $\struct{ A, \tau_A }$ is simply connected.
== Proof ==
Follows from Convex Set is Star Convex Set and Star Convex Set is Simply Connected.
Category:Convex Sets (Vector Spaces)
Category:Simply Connected Spaces
Convex Set is Star Convex Set

== Theorem ==
Let $V$ be a vector space over $\R$ or $\C$.
Let $A \subseteq V$ be a non-empty convex set.
Then $A$ is a star convex set, and every point in $A$ is a star center.
== Proof ==
Let $a \in A$.
Note that there is at least one point in $A$, as $A$ is non-empty.
If $x \in A$, then there is a line segment joining $a$ and $x$.
By definition, it follows that $A$ is star convex, and $a$ is a star center.
Category:Convex Sets (Vector Spaces)
Convex Set of Ordered Set is not necessarily Interval

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $C$ be a convex set of $S$.
Then it is not necessarily the case that $C$ is an interval of $S$.
== Proof ==
Consider the open ray of $S$:
:$R = \set {x \in S: a \prec x}$
for some $a \in S$.
From Ray is Convex, $R$ is a convex set of $S$.
But $R$ is not an interval of $S$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $1$
Category:Convex Sets (Order Theory)
Category:Intervals
Convex Subset of Topological Vector Space containing Zero Vector in Interior is Absorbing Set

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$. 
Let $C \subseteq X$ be a convex set such that ${\mathbf 0}_X \in C^\circ$, where $C^\circ$ denotes the interior of $C$. 
Then $C$ is an absorbing set.
== Proof ==
Let $x \in X$. 
Let $V$ be an open neighborhood of ${\mathbf 0}_X$ contained in $C$. 
Then from Multiple of Vector in Topological Vector Space Converges, we have: 
:$\ds \frac x n \to {\mathbf 0}_X$
So by the definition of a convergent sequence, for some $N \in \N$ we have: 
:$\ds \frac x N \in V$
Then: 
:$\ds x \in N V \subseteq N C$
Since $x \in X$ was arbitrary, we can apply Characterization of Convex Absorbing Set in Vector Space and obtain that: 
:$C$ is absorbing set.
Category:Absorbing Sets
Category:Topological Vector Spaces
Category:Convex Sets (Vector Spaces)