nyuuzyou/wikis huggingface dataset:
Continuous Open Surjective Mapping is Quotient Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: S_1 \to S_2$ be a continuous open surjective mapping.
Then $f$ is a quotient mapping.
== Proof ==
Let $U \subseteq S_2$ such that $f^{-1} \sqbrk U$ is open in $T_1$.
By definition of open mapping:
:$f \sqbrk {f^{-1} \sqbrk U}$ is open in $T_2$
From :
:$f \sqbrk {f^{-1} \sqbrk U} = U$
It follows that $U$ is open in $T_2$.
By definition of quotient mapping, it follows that $f$ is a quotient mapping.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Open Mappings
Category:Quotient Topology
Continuous Real-Valued Function/Examples/Non-Continuous Example 1

== Example of Continuous Real-Valued Function ==
Let $f: \R^2 \to \R$ be the real $2$-variable function defined as:
:$\forall \tuple {x_1, x_2} \in \R^2: \map f {x_1, x_2} = \begin {cases} 0 & : \tuple {x_1, x_2} = \tuple {0, 0} \\ \dfrac {x_1 x_2} {x_1^2 + x_2^2} & : \text {otherwise} \end {cases}$
Then the restrictions of $f$:
:$f_{\restriction \R \times \set 0}$
:$f_{\restriction \set 0 \times \R}$
are both constant functions with value $0$ for all arguments.
Hence both are continuous at $\tuple {0, 0}$.
But $f$ is not continuous at $\tuple {0, 0}$.
== Proof ==
Let $\epsilon = \dfrac 1 2$.
Let $\delta > 0$.
Then:
:the point $x = \tuple {\dfrac \delta 2, \dfrac \delta 2}$ satisfies $\map {d_2} {x, 0} < \delta$
but:
:$\size {\map f x - \map f 0} = \dfrac 1 2$
Thus $f$ is continuous in each variable separately, but not in both variables jointly.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples
Category:Continuous Real-Valued Functions
Continuous Real-Valued Function/Examples/Non-Continuous Example 2

== Example of Continuous Real-Valued Function ==
Let $f: \R^2 \to \R$ be the real $2$-variable function defined as:
:$\forall \tuple {x, y} \in \R^2: \map f {x, y} = \begin {cases} 0 & : y = 0 \\ \dfrac {x^2} y \end {cases}$
Then the restriction of $f$:
:$f_{\restriction \tuple {x, y} \mathop \in \R^2: y \mathop = m x}$
is continuous.
But $f$ is not continuous at $\tuple {0, 0}$.
== Proof ==
Let $N_\epsilon$ be a neighborhood of $\tuple {0, 0}$.
The for sufficiently small $k$, the point $\tuple {k, k^2} \in N_\epsilon$.
But $\map f {k, k^2} = 1$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 24$
Category:Continuous Real-Valued Functions
Continuous Real-Valued Function is not necessarily Bounded

== Theorem ==
Let $S$ be a set.
Let $f: S \to \R$ be a continuous real-valued function.
Then $f$ is not necessarily bounded.
== Proof ==
Let $S$ denote the (open) real interval $\openint 0 1$.
Let $f: S \to \R$ denote the reciprocal function:
:$\forall x \in S: \map f x = \dfrac 1 x$
From Reciprocal Function is Continuous on Real Numbers without Zero, $f$ is continuous on $S$.
From Reciprocal Function is Unbounded on Open Unit Interval, $f$ is unbounded on $S$.
Hence the result, by Proof by Counterexample.
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Example $5.1.2$
Category:Continuous Real-Valued Functions
Category:Bounded Real-Valued Functions
Continuous Real-Valued Function on Compact Space is Bounded

== Theorem ==
Let $\struct {K, \tau}$ be a compact space.
Let $\R$ denote the real number line.
Let $f: S \to \R$ be a continuous real-valued function.
Then:
:$f$ is bounded.
== Proof ==
From Compact Space is Pseudocompact:
:$\struct {K, \tau}$ is pseudocompact
By definition of pseudocompact:
:$f$ is bounded.
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Bounded Real-Valued Functions
Category:Continuous Real-Valued Functions
Category:Pseudocompact Spaces
Category:Compact Spaces
Continuous Real-Valued Function on Irreducible Space is Constant

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Let $f: S \to \R$ be a continuous real-valued function.
Then $f$ is constant, that is:
:$\exists a \in \R: \forall x \in S: \map f x = a$
== Proof ==
By definition of $T = \struct {S, \tau}$ being irreducible:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O\implies U_1 \cap U_2 \ne \O$
For every $x \in \R$, define:
:$L_x := f^{-1} \sqbrk {\openint {-\infty} x}$
:$U_x := f^{-1} \sqbrk {\openint x \infty}$
By continuity of $f$, these are open in $T$.
They are also disjoint, because for each $s \in S$, $\map f s < x$ and $\map f s > x$ cannot both hold simultaneously.
Suppose now that $\map f s = a$ and $\map f {s'} = b$, for some $s, s' \in S$ and $a, b \in \R$ with $a < b$.
Define $c = \dfrac {a + b} 2$.
Then $a < c < b$, whence:
:$s \in L_c$ and $s' \in U_c$
However, because $T$ is irreducible, it follows that either $L_c$ or $U_c$ must be empty.
This contradicts the existence of such $a$ and $b$.
Hence $f$ must be constant.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Irreducible Spaces
Category:Constant Mappings
Continuous Real Function/Examples

== Examples of Continuous Real Functions ==
=== Example: $\sqrt x$ at $x = 1$ ===
=== Example: $\dfrac {\sin x} x$ with $1$ at $x = 0$ ===
=== Example: $\map \sin {\dfrac 1 x}$ with $0$ at $x = 0$ ===
Category:Examples of Continuous Real Functions
Continuous Real Function/Examples/Root of x at 1

== Examples of Continuous Real Functions ==
Let $f: \R_{\ge 0} \to \R$ be the real function defined as:
:$\map f x = \sqrt x$
Then $\map f x$ is continuous at $x = 1$.
== Proof ==
From Limit of Real Function: Example: $\sqrt x$ at $1$, we have that:
The result follows by definition of continuous real function.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity
Category:Examples of Continuous Real Functions
Continuous Real Function/Examples/Sine of Reciprocal of x

== Examples of Continuous Real Functions ==
Let $f: \R_{\ge 0} \to \R$ be the real function defined as:
:$\map f x = \begin {cases} \map \sin {\dfrac 1 x} & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
Then $\map f x$ is not continuous at $x = 0$.
== Proof ==
From Limit of Real Function: $\map \sin {\dfrac 1 x}$ at $0$, $\map f x$ has no limit at $x = 0$.
In order to be continuous at $x = 0$, it needs to have a limit at $x = 0$.
The fact that $\map f x$ is defined at $x = 0$ is immaterial.
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.4$: Continuity: Example $1.4.2$
Category:Examples of Continuous Real Functions
Continuous Real Function/Examples/Sine of x over x with 1 at 0

== Examples of Continuous Real Functions ==
Let $f: \R_{\ge 0} \to \R$ be the real function defined as:
:$\map f x = \begin {cases} \dfrac {\sin x} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Then $\map f x$ is continuous at $x = 0$.
== Proof ==
From Limit of $ \dfrac {\sin x} x$, we have that:
The result follows by definition of continuous real function.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity
Category:Examples of Continuous Real Functions
Continuous Real Function Bounded on Finite Subdivision

== Theorem ==
Let $A = \closedint a b$ be a closed real interval of the set $\R$ of real numbers.
Let $f: A \to \R$ be a continuous real function on $A$.
Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a finite subdivision of $A$ such that:
:each $\closedint {x_j} {x_{j + 1} }$ is a neighborhood of some $a_j$ such that $f$ is bounded on $\closedint {x_j} {x_{j + 1} }$.
Then $f$ is bounded on $A$.
== Proof ==
Follows directly from:
:Continuous Real Function is Bounded on Neighborhood of Argument
and:
:Mapping is Bounded on Union iff Bounded on Each Component/Real-Valued Function.
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Step $4$
Category:Continuous Real-Valued Functions
Category:Boundedness
Continuous Real Function is Baire Function

== Theorem ==
Let $X \subseteq \R$.
Let $f : X \to \R$ be a continuous function.
Then $f$ is a Baire function.
== Proof ==
For each natural number $n$, define: 
:$\map {f_n} x = \map f x$
Since $f$ is continuous: 
:$f_n$ is continuous for each $n$. 
Clearly, for each $x \in X$ we have: 
:$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
from Eventually Constant Sequence Converges to Constant. 
So:
:$\sequence {f_n}$ is a sequence of continuous functions that converges pointwise to $f$.
So $f$ is a Baire function.
Category:Baire Functions
Continuous Real Function is Bounded on Neighborhood of Argument

== Theorem ==
Let $A \subseteq \R$ be a subset of the real number line $\R$.
Let $f: A \to \R$ be a continuous real function on $A$.
Let $a \in A$.
Then there exists a bound:
:$K_a = 1 + \size {\map f a}$ for $\size {\map f x}$
for all $x$ in some neighborhood:
:$\openint {a - \map \delta a} {a + \map \delta a}$ of $a$
where $\map \delta a$ is a positive real constant which depends on $a$.
== Proof ==
Let $a \in A$.
By definition of continuous real function, there exists $\delta \in \R_{>0}$ such that:
:$\forall x \in A: 0 < \size {x - a} < \delta \implies \size {\map f x - \map f a} < \epsilon$
for all $\epsilon \in \R_{>0}$.
Putting $\epsilon = 1$, say, gives us:
:$\forall x \in A: 0 < \size {x - a} < \delta \implies \size {\map f x - \map f a} < 1$
Hence:
:$\size {\map f x} < 1 + \size {\map f a}$
for all $x \in \openint {a - \delta} {a + \delta}$
Here it is important to note that $\delta$ is dependent upon both $a$ and $f$, as well as $\epsilon$ (chosen here to be $1$).
== Sources ==
* : $5$: Compact spaces: $5.1$: Motivation: Step $3$
Category:Continuous Real-Valued Functions
Category:Bounded Real-Valued Functions
Continuous Real Function is Darboux Integrable

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.
== Proof ==
We have the result $f$ is bounded by Continuous Real Function is Bounded.
By Condition for Darboux Integrability, it suffices to show that for all $\epsilon > 0$, there exists a subdivision $P$ of $\closedint a b$ such that:
:$\map U P – \map L P < \epsilon$
where $\map U P$ and $\map L P$ denote the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ belonging to the subdivision $P$.
Let $\epsilon > 0$.
We have the result Continuous Function on Closed Real Interval is Uniformly Continuous.
By the definition of uniform continuity, there exists a $\delta > 0$ such that if $x, y \in \closedint a b$ are such that $\size {x – y} < \delta$, then:
:$\size {\map f x – \map f y} < \dfrac \epsilon {b - a}$
Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a subdivision of $\closedint a b$ such that:
:$\ds \max_{1 \mathop \le k \mathop \le n} \paren {x_k – x_{k - 1} } < \delta$
For all integers $k$ satisfying $1 \le k \le n$, it follows from the Heine-Borel theorem that $\closedint {x_{k - 1} } {x_k}$ is compact.
So we can apply Corollary 3 to Continuous Image of Compact Space is Compact to conclude that there exist $u_k, v_k \in \closedint {x_{k - 1} } {x_k}$ such that:
By assumption, $x_k – x_{k - 1} < \delta$, so:
:$\size {u_k – v_k} < \delta$
It follows from the definition of $\delta$ that:
:$\ds \map f {u_k} – \map f {v_k} < \frac \epsilon {b – a}$
This gives:
as desired.
== Notes ==
If $f$ is not continuous on $\closedint a b$ (while still bounded), then the lower and upper integral do not necessarily have the same result. But they might.
== Sources ==
* : $6.8$
* : Chapter $1$: Fourier Series: $1.1$ Basic Concepts: $1.1.3$ Definitions
* : $2.5$: The Riemann Integral
Category:Integral Calculus
Category:Continuous Real Functions
Continuous Real Function on Closed Interval/Examples/Reciprocal of 1 + e to the Reciprocal of x

== Examples of Continuous Real Functions on Closed Intervals ==
Consider the real function $f$ defined as:
:$f := \begin {cases} \dfrac 1 {1 + e^{1 / x} } & : x \ne 0 \\ 0 & : x = 0 \end {cases}$
Then $f$ is continuous on the closed interval $\closedint 0 1$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity
Category:Examples of Continuous Real Functions
Continuous Real Function on Closed Interval is Bijective iff Strictly Monotone

== Theorem ==
Let $\closedint a b$ and $\closedint c d$ be closed real intervals.
Let $f: \closedint c d \to \closedint a b$ be a continuous real function.
Let $\map f c, \map f d \in \set {a, b}$.
Then $f$ is bijective $f$ is strictly monotone.
== Proof ==
=== Necessary condition ===
Let $f$ be a bijection.
From Continuous Injection of Interval is Strictly Monotone, it follows that $f$ is strictly monotone.
=== Sufficient condition ===
Let $f$ be strictly monotone.
From Strictly Monotone Real Function is Bijective, it follows that $f$ is a bijection on its image.
From Image of Interval by Continuous Function is Interval, it follows that the image of $f$ is a real interval.
Suppose that $f$ is strictly increasing.
Then by definition $\map f c < \map f d$.
It follows that:
It follows that for all $x \in \closedint c d$:
Thus, the image of $f$ is $\closedint a b$.
Suppose instead that $f$ is strictly decreasing.
Then by definition $\map f c > \map f d$.
It follows that:
It follows that for all $x \in \closedint c d$:
Thus the image of $f$ is again $\closedint a b$.
Category:Continuous Real Functions
Category:Monotone Real Functions
Category:Bijections
Continuous Real Linear Functional on Complex Topological Vector Space is Real Part of Continuous Complex Linear Functional

== Theorem ==
Let $\struct {X, \tau}$ be a topological vector space over $\C$. 
Let $g : X \to \R$ be a continuous continuous $\R$-linear functional.
Then there exists a unique continuous continuous $\C$-linear functional $f : X \to \C$ such that: 
:$\map g x = \map \Re {\map f x}$ 
for each $x \in X$. 
== Proof ==
From Real Linear Functional is Real Part of Unique Linear Functional, there exists a unique $\C$-linear functional $f : X \to \C$ such that: 
:$\map g x = \map \Re {\map f x}$ 
for each $x \in X$, given by: 
:$\map f x = \map g x - i \map g {i x}$
We need to show that $g$ is continuous.
From the definition of a topological vector space, the map $c : \C \times X \to X$ defined by: 
:$\map c {\lambda, x} = \lambda x$
for each $\tuple {\lambda, x} \in \C \times X$ is continuous.
From Vertical Section of Continuous Function is Continuous, the map $c_i : X \to X$ defined by: 
:$\map {c_i} x = i x$
for each $x \in X$ is continuous.
From Composite of Continuous Mappings is Continuous:
:$x \mapsto \map g {i x}$ is continuous.
From Linear Combination of Continuous Functions valued in Topological Vector Space is Continuous, we have that $f$ is continuous.
Category:Topological Vector Spaces
Category:Linear Functionals
Continuous Replicative Function

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $f$ be continuous on $\R$.
Let $f$ also be a replicative function.
Then $f$ is of the form:
:$\map f x = \paren {x - \dfrac 1 2} a$
where $a \in \R$.
== Proof ==
Let $f$ be a replicative function.
Then:
:$\forall n > 0: \map f {n x + 1} - \map f {n x} = \map f {x + 1} - \map f x$
If $f$ is then also continuous:
:$\forall x \in \R: \map f {x + 1} - \map f x$
and so:
:$\map g x = \map f x - c \floor x$
is both replicative and periodic.
We have:
:$\ds \int_0^1 e^{2 \pi i n x} \map g x \rd x = \dfrac 1 n \int_0^1 e^{2 \pi y} \map g y \rd y$
Expanding in a Fourier series shows:
:$\map g x = \paren {x - \dfrac 1 2} a$
for $0 < x < 1$.
Thus it follows that:
:$\map f x = \paren {x - \dfrac 1 2} a$
== Historical Note ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $40$
Category:Replicative Functions
Category:Continuous Functions
Continuous Strictly Midpoint-Concave Function is Strictly Concave

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
Let $f$ be strictly midpoint-concave and continuous on $I$.
Then $f$ is strictly concave.
== Proof ==
== Also see ==
* Continuous Midpoint-Concave Function is Concave
* Continuous Midpoint-Convex Function is Convex
* Continuous Strictly Midpoint-Convex Function is Strictly Convex
Category:Real Analysis
Continuous Strictly Midpoint-Convex Function is Strictly Convex

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
Let $f$ be strictly midpoint-convex and continuous on $I$.
Then $f$ is strictly convex.
== Proof ==
== Also see ==
* Continuous Midpoint-Convex Function is Convex
* Continuous Midpoint-Concave Function is Concave
* Continuous Strictly Midpoint-Concave Function is Strictly Concave
Category: Real Analysis
Continuous Surjection Induces Continuous Bijection from Quotient Space

== Theorem ==
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $g: S_1 \to S_2$ be a $\tuple {\tau_1, \tau_2}$-continuous surjection.
Let $\RR_g \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $g$:
:$\tuple {s_1, s_2} \in \RR_g \iff \map g {s_1} = \map g {s_2}$
Let $q_{\RR_g}: S_1 \to S_1 / \RR_g$ be the canonical surjection induced by the equivalence relation $\RR_g$.
Let $\tau_{\RR_g}$ be the quotient topology on the quotient space $S_1 / \RR_g$ by $q_{\RR_g}$.
Then $g$ induces a $\tuple {\tau_{\RR_g}, \tau_2}$-continuous bijection $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.
$\qquad \begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{ q_{\RR_g} }
 \ar[rd]_*{g}
&
 S_1 / \RR_g \ar@{-->}[d]^*{f}
\\ &
 S_2
}\end{xy}$
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
By definition of quotient topology, $q_{\RR_g}$ is a surjective identification mapping.
Quotient Mapping equals Surjective Identification Mapping shows that $q_{\RR_g}$ is a quotient mapping.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping shows that $g$ induces a continuous mapping $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.
As $g$ is surjective, it follows that $f$ is surjective.
Let $\eqclass {s_1} {\RR_g}, \eqclass {s_2} {\RR_g} \in S_1 / \RR_g$ such that $\map f {\eqclass {s_1} {\RR_g} } = \map f {\eqclass {s_2}{ \RR_g} }$.
As $f \circ q_{\RR_g} = g$, it follows that $\map g {s_1} = \map g {s_2}$.
By definition of equivalence relation induced by $g$, it follows that $\eqclass {s_1} {\RR_g} = \eqclass {s_2} {\RR_g}$.
By definition of injection, it follows that $f$ is injective.
As $f$ is injective and surjective, it follows that $f$ is bijective.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Spaces
Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 1

== Theorem ==
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $g: S_1 \to S_2$ be a $\tuple {\tau_1, \tau_2}$-continuous surjection.
Let $\RR_g \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $g$:
:$\tuple {s_1, s_2} \in \RR_g \iff \map g {s_1} = \map g {s_2}$
Let $q_{\RR_g}: S_1 \to S_1 / \RR_g$ be the canonical surjection induced by the equivalence relation $\RR_g$.
Let $\tau_{\RR_g}$ be the quotient topology on the quotient space $S_1 / \RR_g$ by $q_{\RR_g}$.
Then $g$ induces a continuous bijection $f: S_1 / \RR_g \to S_2$ such that $f \circ q_{\RR_g} = g$.
The induced mapping $f$ is a homeomorphism, $g$ is a quotient mapping.
$\qquad \begin{xy} \xymatrix@L+2mu@+1em{
 S_1 \ar[r]^*{ q_{\RR_g} }
 \ar[rd]_*{g}
&
 S_1 / \RR_g \ar@{-->}[d]^*{f}
\\ &
 S_2
}\end{xy}$
== Proof ==
Continuous Surjection Induces Continuous Bijection from Quotient Space shows that $g$ is well-defined.
Let $f$ be a homeomorphism.
Injective Quotient Mapping Equals Homeomorphism shows that $f$ is a quotient mapping.
Composite of Quotient Mappings in Topology is Quotient Mapping shows that $g = f \circ q_{\RR_g}$ is a quotient mapping.
Let $g$ be a quotient mapping.
Quotient Mapping and Continuous Mapping Induces Continuous Mapping/Corollary shows that $f$ is a quotient mapping.
Injective Quotient Mapping Equals Homeomorphism shows that $f$ is a homeomorphism.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Spaces
Continuous Surjection Induces Continuous Bijection from Quotient Space/Corollary 2

== Theorem ==
Let $\struct {S_1, \tau_1}$ and $\struct {S_2, \tau_2}$ be topological spaces.
Let $g: S_1 \to S_2$ be a continuous surjection.
Let $\RR_g \subseteq S_1 \times S_1$ be the equivalence on $S_1$ induced by $g$:
:$\tuple {s_1, s_2} \in \RR_g \iff \map g {s_1} = \map g {s_2}$
Let $q_{\RR_g}: S_1 \to S_1 / \RR_g$ be the canonical surjection induced by the equivalence relation $\RR_g$.
Let $\tau_{\RR_g}$ be the quotient topology on the quotient space $S_1 / \RR_g$ by $q_{\RR_g}$.
If $S_2$ is a Hausdorff space, then $S_1 / \RR_g$ is a Hausdorff space.
== Proof ==
Continuous Surjection Induces Continuous Bijection from Quotient Space shows there exists a continuous bijection $f: S_1 / \RR_g \to S_2$.
Let $\eqclass {s_1}{\RR_g}, \eqclass {s_2}{\RR_g} \in S_1 / \RR_g$ such that $\eqclass {s_1}{\RR_g} \ne \eqclass {s_2}{\RR_g}$.
As $f$ is injective, $\map f { \eqclass {s_1}{\RR_g} } \ne \map f { \eqclass {s_2}{\RR_g} }$.
As $S_2$ is a Hausdorff space, there exists disjoint open sets $U_1, U_2 \in \tau_2$ such that $\map f { \eqclass {s_1}{\RR_g} } \in U_1$ and $\map f { \eqclass {s_2}{\RR_g} } \in U_2$.
It follows that $f^{-1} \sqbrk {U_1}$ and $f^{-1} \sqbrk {U_2}$ are disjoint.
As $f$ is continuous, we have $f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2} \in \tau_{\RR_g}$.
By definition of Hausdorff space, $S_1 / \RR_g$ is a Hausdorff space.
== Sources ==
* : $2$: Topological Spaces and Continuous Functions: $\S 22$: The Quotient Topology
Category:Quotient Spaces
Category:Hausdorff Spaces
Continuous Variable/Examples/Age of a Person

== Example of Continuous Variable ==
The age $A$ of a given person can take any of the values in the set $\R_{\ge 0} = \set {x \in \R: x \ge 0}$.
Thus $A$ is a continuous variable.
== Sources ==
* : Chapter $1$: Discrete and Continuous Variables: Example 2.
Category:Examples of Continuous Variables
Continuous at Point iff Left-Continuous and Right-Continuous

== Theorem ==
Let $A \subseteq \R$ be an open set of real numbers.
Let $f : A \to \R$ be a real function.
Let $x_0 \in A$.
Then:
:$f$ is continuous at $x_0$
:
:$f$ is both left-continuous and right-continuous at $x_0$
== Proof ==
=== Necessary Condition ===
Suppose $f$ is continuous at $x_0$.
Then, by definition:
:$\ds \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$
By Limit iff Limits from Left and Right:
:$\ds \lim_{x \mathop \to x_0^-} \map f x = \map f {x_0}$
:$\ds \lim_{x \mathop \to x_0^+} \map f x = \map f {x_0}$
Therefore, by definition, $f$ is both left-continuous and right-continuous at $x_0$.
=== Sufficient Condition ===
Suppose $f$ is both left-continuous and right-continuous at $x_0$.
Then, by the respective definitions:
:$\ds \lim_{x \mathop \to x_0^-} \map f x = \map f {x_0}$
:$\ds \lim_{x \mathop \to x_0^+} \map f x = \map f {x_0}$
By Limit iff Limits from Left and Right:
:$\ds \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$
Therefore, by definition, $f$ is continuous at $x_0$.
Category:Continuous Real Functions
Continuous iff Directed Suprema Preserving

== Theorem ==
Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be complete topological lattices with Scott topologies.
Let $f: S \to T$ be a mapping.
Then $f$ is continuous $f$ preserves directed suprema.
== Proof ==
=== Sufficient Condition ===
Assume that
:$f$ is continuous.
By Continuous iff Mapping at Limit Inferior Precedes Limit Inferior of Composition of Mapping and Sequence:
:for all directed set $\struct {D, \precsim}$ and net $N:D \to S$ in $S$: $\map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$
Thus by Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema:
:$f$ preserves directed suprema.
=== Necessary Condition ===
Thus by Mapping Preserves Directed Suprema implies Mapping is Continuous:
:if $f$ preserves directed suprema, then $f$ is continuous.
== Sources ==
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Category:Topological Order Theory