nyuuzyou/wikis huggingface dataset:
Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set/Lemma 1

== Lemma ==
Let $\struct {X, \tau}$ be a compact Hausdorff space. 
Let $f : X \to X$ be a continuous function.
Define a sequence of sets $\sequence {X_i}_{i \mathop \in \N}$ by:
:$X_i = \begin{cases} X & : i = 1 \\ \map f {X_{i - 1} } & : i \ge 2 \end{cases}$
Then, for each $i \in \N$:
:$X_i$ is closed
:$X_i$ is non-empty
:$X_{i + 1} \subseteq X_i$.
== Proof ==
For all $n \in \N$, let $\map P n$ be the proposition:
:the set $X_n$ is non-empty and closed with $X_{n + 1} \subseteq X_n$.
=== Base Case ===
From Underlying Set of Topological Space is Clopen:
:$X$ is closed. 
We have that $X \ne \O$.
We also have that: 
:$X_2 = \map f {X_1} \subseteq X = X_1$
So $\map P 1$ is true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So, suppose that: 
:$X_k$ is closed and non-empty,
and:
:$X_{k + 1} \subseteq X_k$
This is our induction hypothesis.
=== Induction Step ===
Now we need to show that: 
:$X_{k + 1}$ is closed and non-empty,
and:
:$X_{k + 2} \subseteq X_{k + 1}$
We have: 
:$X_{k + 1} = \map f {X_k}$
Note that: 
:$X_k \subseteq X$
Since $X_k$ is non-empty, we have: 
:$X_{k + 1} \ne \O$
From Closed Subspace of Compact Space is Compact, we have: 
:$X_k$ is compact. 
From Continuous Image of Compact Space is Compact, we therefore have: 
:$X_{k + 1} = \map f {X_k}$ is compact.
From Compact Subspace of Hausdorff Space is Closed, we have: 
:$X_{k + 1}$ is closed.
So:
:$X_{k + 1}$ is non-empty and closed.
From: 
:$X_{k + 1} \subseteq X_k$
We obtain: 
:$\map f {X_{k + 1} } \subseteq \map f {X_k}$
and so:
:$X_{k + 2} \subseteq X_{k + 1}$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Category:Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set
Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set/Lemma 2

== Lemma ==
Let $\struct {T, \tau}$ be a compact Hausdorff space. 
Let $f : T \to T$ be a continuous function.
Define a sequence of sets $\sequence {X_i}_{i \mathop \in \N}$ by:
:$X_i = \begin{cases}X & : i = 1 \\ f \sqbrk {X_{i - 1} } & : i \ge 2\end{cases}$
Define:
:$\ds A = \bigcap_{n \mathop = 1}^\infty X_n$
Then: 
:$A \subseteq f \sqbrk A$
== Proof ==
Let $y \in A$. 
Then, by the definition of set intersection:
:$y \in f \sqbrk {X_i}$ for each $i \in \N$.
That is:
:for each $i \in \N$, there exists $x_i \in X_i$ such that $\map f {x_i} = y$. 
So the intersection $f^{-1} \sqbrk {\set y} \cap X_i$ is non-empty for each $i \in \N$.
From Continuity Defined from Closed Sets:
:$f^{-1} \sqbrk {\set y}$ is closed.
From Intersection of Closed Sets is Closed in Topological Space:
:$f^{-1} \sqbrk {\set y} \cap X_i$ is closed.
From Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set: Lemma 1, we have: 
:$X_{i + 1} \subseteq X_i$ for each $i \in \N$. 
So, from Set Intersection Preserves Subsets: 
:$f^{-1} \sqbrk {\set y} \cap X_{i + 1} \subseteq {f^{-1} } \sqbrk {\set y} \cap X_i$
We can therefore apply Intersection of Nested Closed Subsets of Compact Space is Non-Empty to obtain: 
:$\ds \bigcap_{i \mathop = 1}^\infty \paren {f^{-1} \sqbrk {\set y} \cap X_i} \ne \O$
We then have: 
So there exists $x \in A$ such that: 
:$\map f x = y$
That is: 
:$y \in f \sqbrk A$
So we obtain: 
:$A \subseteq f \sqbrk A$
as required.
Category:Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set
Continuous Function is not necessarily Differentiable

== Theorem ==
Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is continuous at $x_0$.
Then it is not necessarily the case that $f$ is differentiable at $x_0$.
== Proof ==
;Proof by Counterexample
Consider the real function $f: \R \to \R$ defined as:
:$\forall x \in \R: \map f x = \size x$
where $\size x$ denotes the absolute value function.
From Absolute Value Function is Continuous, $f$ is continuous on $\R$.
From Derivative of Absolute Value Function, $f$ is not differentiable at $x = 0$.
Hence $f$ is an example of a real function which at $x_0 = 0$ is continuous at $x_0$ but not differentiable at $x_0$.
== Also see ==
* Differentiable Function is Continuous
== Sources ==
* 
* 
Category:Differentiable Real Functions
Category:Continuous Real Functions
Continuous Function on Closed Real Interval is Uniformly Continuous

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $f: \closedint a b \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $\closedint a b$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* 
* 
Category:Continuous Function on Closed Real Interval is Uniformly Continuous
Category:Continuous Real Functions
Category:Uniformly Continuous Real Functions
Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 1

== Theorem ==
== Proof ==
We have that $\R$ is a metric space under the usual (Euclidean) metric.
We also have from the Heine-Borel Theorem that $\closedint a b$ is compact.
So the result Heine-Cantor Theorem applies.
== Sources ==
* : $5$: Compact spaces: $5.8$: Compactness and Uniform Continuity
Category:Continuous Function on Closed Real Interval is Uniformly Continuous
Continuous Function on Closed Real Interval is Uniformly Continuous/Proof 2

== Theorem ==
== Proof ==
 $f$ is not uniformly continuous. 
Then there exists an $\epsilon \in \R_{>0}$ such that for all $\delta \in \R_{>0}$ there are points $x ,y \in \closedint a b$ for which:
:$\size {x - y} < \delta$
and:
:$\size {\map f x - \map f y} \ge \epsilon$
We will now choose for each $k \in \N$ numbers $x_k, y_k \in \closedint a b$ such that:
:$\size {x_k - y_k} < \dfrac 1 k$
and:
:$\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$
We have that the sequence $\sequence {x_k}$ is bounded in $\closedint a b$.
So, by the Bolzano-Weierstrass Theorem there exists a convergent subsequence $\sequence {x_{k_j} }$ whose limit lies in $\closedint a b$.
Let this limit be denoted $x_0$.
Now we have:
:$\size {x_0 - y_{k_j} } \le \size {x_0 - x_{k_j} } + \size {x_{k_j} - y_{k_j} } \le \size {x_0 - x_{k_j} } + \dfrac 1 {k_j}$
Therefore the sequence $\sequence {y_{k_j} }$ also converges to $x_0$.
Because $f$ is sequentially continuous at $x_0 \in \closedint a b$, we have:
:$\ds \lim_{j \mathop \to \infty} \map f {x_{k_j} } = \map f {x_0} = \lim_{j \mathop \to \infty} \map f {y_{k_j} }$
This is however, a contradiction, as $\size {\map f {x_k} - \map f {y_k} } \ge \epsilon$ for all $k$ and thus all $k_j$. 
Therefore $f$ is uniformly continuous.
Category:Continuous Function on Closed Real Interval is Uniformly Continuous
Continuous Function on Compact Space is Bounded

== Theorem ==
Let $\struct {X, \tau}$ be a compact topological space.
Let $\struct {Y, \norm {\, \cdot \, } }$ be a normed vector space.
Let $f: X \to Y$ be continuous.
Then $f$ is bounded.
== Proof 1 ==
 
== Proof 2 ==
Category:Continuous Functions
Category:Compact Spaces
Category:Continuous Function on Compact Space is Bounded
Continuous Function on Compact Space is Bounded/Proof 1

== Theorem ==
== Proof ==
 $f$ is not bounded.
Let $g : X \to \R$ be defined by:
:$\map g x = \norm {\map f x}$
From Norm is Continuous and Composite of Continuous Mappings is Continuous, it follows that $g$ is continuous.
For all $n \in \N$, set $A_n := g^{-1} \sqbrk {\map {B_n} 0}$, where $\map {B_n} 0$ denotes the open ball in $\R$ with radius $n$ and center $0$.
From Open Ball is Open Set in Normed Vector Space and the definition of continuity, it follows that all $A_n$ are open in $X$.
For all $x \in X$, we have:
:$x \in A_n$ for all $n \ge \norm {\map f x}$
It follows that:
:$\ds X \subseteq \bigcup_{n \mathop \in \N} A_n$
By definition of open cover, $\set {A_n}_{n \mathop \in \N}$ is an open cover of $X$.
As $X$ is compact, there exist $m \in \N$ and $n_1, \dots, n_m \in \N$ such that $\set {A_{n_k} }_{k \mathop \in \set {1, \ldots, m} }$ is a finite subcover of $X$.
It follows that:
:$\ds X \subseteq \bigcup_{k \mathop \in \set {1, \ldots, m} } A_{n_k}$
As $f$ is not bounded, there exists $x_0 \in X$ such that:
:$\norm {\map f {x_0} } > \max \set {n_1, \ldots, n_m}$
It follows that:
:$x_0 \notin \ds \bigcup_{k \mathop \in \set {1, \ldots, m} } A_{n_k}$
which is a contradiction.
Category:Continuous Function on Compact Space is Bounded
Continuous Function on Compact Space is Bounded/Proof 2

== Theorem ==
== Proof ==
From Continuous Image of Compact Space is Compact, $f \sqbrk X$ is a compact subset of $Y$. 
From Compact Subset of Normed Vector Space is Closed and Bounded, $f \sqbrk X$ is bounded.
Hence there exists a real number $M > 0$ such that:
:$\norm {\map f x} \le M$ for all $x \in X$. 
So $f$ is bounded.
Category:Continuous Function on Compact Space is Bounded
Continuous Function on Compact Space is Uniformly Continuous

== Theorem ==
Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $S \subseteq \R^n$ be a compact subspace of $\R^n$.
Let $f: S \to \R$ be a continuous function.
Then $f$ is uniformly continuous on $S$.
== Proof ==
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Continuous Functions
Category:Uniformly Continuous Functions
Continuous Function on Compact Subspace of Euclidean Space is Bounded

== Theorem ==
Let $\R^n$ be the $n$-dimensional Euclidean space.
Let $S \subseteq \R^n$ be a compact subspace of $\R^n$.
Let $f: S \to \R$ be a continuous function.
Then $f$ is bounded in $\R$.
== Proof ==
An application of Continuous Function on Compact Space is Bounded.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Analysis
Category:Continuity
Continuous Functional Calculus Commutes with Character on Generated C*-Subalgebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $\map \CC {\map {\sigma_A} x}$ be the space of continuous functions on $\map {\sigma_A} x$.
Let $f \in \map \CC {\map {\sigma_A} x}$.
Let $\map f x$ be the continuous functional calculus of $x$ applied to $f$.
Let $B$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$. 
Let $\Phi_B$ be the spectrum of $B$. 
Let $\phi \in \Phi_B$.
Then $\map \phi {\map f x} = \map f {\map \phi x}$. 
== Proof ==
From Spectrum of Element of Unital C*-Subalgebra of Unital C*-Algebra, we have:
:$\map {\sigma_A} {\map f x} = \map {\sigma_B} {\map f x}$
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $\map {\sigma_B} x$. 
From Subalgebra Generated by Inclusion and Conjugate is Everywhere Dense in Space of Continuous Functions on Compact Subset of Complex Numbers, there exists a sequence $\sequence {p_k}_{k \in \N}$ such that:
:$\ds \map {p_k} z = \sum_{j \mathop = 1}^{N_k} \alpha_{j, k} z^{n_{j, k} } \overline z^{m_{j, k} }$
for each $z \in \map {\sigma_B} x$ and some $N_k \in \N$, $\alpha_{j, k} \in \C$, $n_{j, k}, m_{j, k} \ge 0$, such that:
:$\norm {p_k - f}_\infty \to 0$ as $n \to \infty$
From Existence and Uniqueness of Continuous Functional Calculus, we have that $\Theta_x$ is a linear isometry and so:
:$\norm {\map {p_k} x - \map f x} \to 0$ as $n \to \infty$
Since $\Theta_x$ is an algebra homomorphism, we note that:
From Character on Banach Algebra is Continuous, we have that $\phi$ is bounded, giving:
:$\cmod {\map \phi {\map {p_k} x} - \map \phi {\map f x} } \to 0$ as $n \to \infty$
From Character on C*-Algebra is *-Algebra Homomorphism, $\phi$ is a $\ast$-algebra homomorphism. 
Hence we have:
:$\ds \map \phi {\map {p_k} x} = \sum_{j \mathop = 1}^{N_k} \alpha_{j, k} \map \phi { x^{n_{j, k} } \paren {x^\ast}^{m_{j, k} } } = \sum_{j \mathop = 1}^{N_k} \alpha_{j, k} \paren {\map \phi x}^{n_{j, k} } \overline {\map \phi x}^{m_{j, k} } = \map {p_k} {\map \phi x}$
Moreover since $\norm {p_k - f}_\infty \to 0$ as $n \to \infty$, we obtain $\map {p_k} {\map \phi x} \to \map f {\map \phi x}$ as $n \to \infty$.
We have already established that $\map {p_k} {\map \phi x} = \map \phi {\map {p_k} x} \to \map \phi {\map f x}$ as $n \to \infty$.
Hence we conclude that $\map f {\map \phi x} = \map \phi {\map f x}$.
Category:Continuous Functional Calculus
Category:Characters (Banach Algebras)
Continuous Functional Calculus obeys Composition

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal. 
Let $\map {\sigma_A} x$ denote the spectrum of $x$ in $A$.
Let $\map \CC {\map {\sigma_A} x}$ be the space of continuous functions on $\map {\sigma_A} x$.
Let $f \in \map \CC {\map {\sigma_A} x}$.
Let $g \in \map \CC {f \sqbrk {\map {\sigma_A} x} }$.
Let $\map {\Theta_x} {g \circ f} = \map {\paren {g \circ f} } x$ be the continuous functional calculus of $x$ applied to $g \circ f$.
Let $\map {\Theta_{\map f x} } g = \map g {\map f x}$ be the continuous functional calculus of $\map f x$ applied to $g$.
Then $\map {\paren {g \circ f} } x = \map g {\map f x}$.
== Proof ==
Note from Existence and Uniqueness of Continuous Functional Calculus that $\map f x$ is normal.
From the Spectral Mapping Theorem, we have:
:$f \sqbrk {\map {\sigma_A} x} = \map {\sigma_A} {\map f x}$
and hence $g \in \map \CC {\map {\sigma_A} {\map f x} }$.
We are hence assured that $\map g {\map f x}$ is well-defined.
Let $B \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, x, x^\ast}$.
Let $C \subseteq A$ be the $\text C^\ast$-algebra generated by $\set { {\mathbf 1}_A, \map f x, \map f x^\ast}$.
From Existence and Uniqueness of Continuous Functional Calculus, we have $C \subseteq B$. 
By C*-Algebra Generated by Commutative Self-Adjoint Set is Commutative, $B$ and $C$ are commutative.
Let $\Phi_B$ and $\Phi_C$ be the spectrums of $B$ and $C$ respectively. 
Let $\phi \in \Phi_B$ so that $\phi \restriction C \in \Phi_C$.
We then have:
Hence we have:
:$\widehat {\map {\paren {g \circ f} } x} = \widehat {\map g {\map f x} }$
where $\hat \cdot$ is the Gelfand transform.
From Gelfand-Naimark Theorem: Commutative Case, we therefore have:
:$\map {\paren {g \circ f} } x = \map g {\map f x}$
Category:Continuous Functional Calculus
Continuous Functional Calculus of Reciprocal is Inverse Element

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $x \in A$ be normal and invertible.
Let $\map {\sigma_A} x$ be the spectrum of $x$ in $A$.
Let $\Theta_x$ be the continuous functional calculus for $x$.
Let $\iota : \map {\sigma_A} x \to \C$ be the inclusion.
Define $1/\iota : \map {\sigma_A} x \to \C$ by:
:$\map {\paren {1/\iota} } z = \dfrac 1 z$
for each $z \in \map {\sigma_A} x$.
Then:
:$\map {\Theta_x} {1/\iota} = x^{-1}$
== Proof ==
Since $x$ is invertible, we have that $0 \not \in \map {\sigma_A} x$ and indeed $1/\iota$ is well-defined.
We have:
and:
Hence we have:
:$\map {\Theta_x} {1/\iota} = x^{-1}$
Category:Continuous Functional Calculus
Continuous Functions on Compact Space form Banach Space

== Theorem ==
Let $X$ be a compact Hausdorff space.
Let $Y$ be a Banach space.
Let $\CC = \CC \struct {X; Y}$ be the space of continuous functions on $X$ valued in $Y$. 
Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on $\CC$.
Then $\struct {\CC, \norm {\,\cdot\,}_\infty}$ is a Banach space.
== Proof ==
From Continuous Function on Compact Space is Bounded, every continuous mapping $f : X \to Y$ is bounded.
Hence $\map \CC {X; Y} = \map {\CC_b} {X; Y}$, where $\map {\CC_b} {X; Y}$ is the space of bounded continuous functions on $X$ valued in $Y$.
From Bounded Continuous Functions on Topological Space form Banach Space, we have that $\struct {\map {\CC_b} {X; Y}, \norm {\, \cdot \,}_\infty} = \struct {\CC, \norm {\,\cdot\,}_\infty}$ is a Banach space.
Category:Functional Analysis
Continuous Functions with Compact Support Dense in Lebesgue P-Space

== Theorem ==
Let $\map {C_c} {\R^n}$ be the space of continuous functions with compact support on $\R^n$.
Let $p \in \R$ such that $p \ge 1$.
Let $\map {\LL^p} {\lambda^n}$ be the Lebesgue $p$-space for Lebesgue measure $\lambda^n$.
Then $\map {C_c} {\R^n}$ is everywhere dense in $\map {\LL^p} {\lambda^n}$ with respect to the $p$-seminorm $\norm {\, \cdot \,}_p$.
== Proof ==
== Sources ==
* : $14.7$
Category:Lebesgue Spaces
Continuous Group Action is by Homeomorphisms

== Theorem ==
Let $G$ be a topological group acting continuously on a topological space $X$.
Then $G$ acts by homeomorphisms.
== Proof ==
Let $\phi: G \times X \to X$ denote the group action.
Let $g \in G$.
We are given that $\phi$ is a bijection.
The map $\phi_g: X \to X : x \mapsto \map \phi {g, x}$ is continuous because $\phi$ is.
By Inverse of Bijection is Bijection, the inverse of $\phi$ is given by $\phi_{g^{-1} }$ is also a bijection.
So the inverse of $\phi$ is continuous.
So inverse of $\phi_g$ is continuous.
Thus $\phi_g$ is an homeomorphism of $X$.
== Also see ==
* Discrete Group Acts Continuously iff Acts by Homeomorphisms
Category:Topological Group Actions
Continuous Image of Closed Interval is Closed Interval

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.
Then the image of $\left[{a \,.\,.\, b}\right]$ under $f$ is also a closed interval.
== Proof ==
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.
== Also known as ==
Some sources refer to this as the continuity property, but this is not standard, and is too easily confused with the Continuum Property.
== Also see ==
* Extreme Value Theorem
== Sources ==
* : $\S 9.8$
Category:Real Analysis
Category:Continuous Functions
Continuous Image of Compact Space is Compact

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $f: T_1 \to T_2$ be a continuous mapping.
If $T_1$ is compact then so is its image $f \sqbrk {T_1}$ under $f$.
That is, compactness is a continuous invariant.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
* 
Category:Compact Spaces
Category:Continuous Invariants
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Corollary 1

== Corollary to Continuous Image of Compact Space is Compact ==
Compactness is a topological property.
== Proof ==
Follows directly from:
:Continuous Image of Compact Space is Compact
:the definition of topological property
:the definition of homeomorphism.
== Sources ==
* : $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Corollary $5.5.2$
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Corollary 2

== Corollary to Continuous Image of Compact Space is Compact ==
A continuous mapping from a compact topological space to a metric space is bounded.
== Proof ==
Follows from Continuous Image of Compact Space is Compact and Compact Subspace of Metric Space is Bounded.
== Sources ==
* : $4.15$
* : $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Corollary $5.5.3$
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Corollary 3

== Corollary to Continuous Image of Compact Space is Compact ==
Let $S$ be a compact topological space.
Let $f: S \to \R$ be a continuous real-valued function.
Then $f$ attains its bounds on $S$.
== Proof 1 ==
== Proof 2 ==
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Corollary 3/Proof 1

== Corollary to Continuous Image of Compact Space is Compact ==
== Proof ==
By , $f \sqbrk S$ is bounded.
By Supremum of Bounded Above Set of Reals is in Closure:
:$\map \sup {f \sqbrk S} \in \map \cl {f \sqbrk S}$
and by Infimum of Bounded Below Set of Reals is in Closure:
:$\map \inf {f \sqbrk S} \in \map \cl {f \sqbrk S}$
From Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact in $\R$.
From Non-Closed Set of Real Numbers is not Compact, it follows from the Rule of Transposition that $f \sqbrk S$ is closed in $\R$.
From Closed Set equals its Closure:
:$f \sqbrk S = \map \cl {f \sqbrk S}$
Hence the result that:
:$\map \sup {f \sqbrk S} \in f \sqbrk S$
and:
:$\map \inf {f \sqbrk S} \in f \sqbrk S$
== Sources ==
* : $4.16$
* : $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Corollary $5.5.4$
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Corollary 3/Proof 2

== Corollary to Continuous Image of Compact Space is Compact ==
== Proof ==
By Continuous Image of Compact Space is Compact, $f \sqbrk S$ is compact.
From Compact Metric Space is Complete and Compact Metric Space is Totally Bounded, $f \sqbrk S$ is complete and totally bounded.
A Totally Bounded Metric Space is Bounded.
Hence both the supremum and the infimum of $f \sqbrk S$ exist in $\R$.
Because $f \sqbrk S$ is complete:
:$\sup f \sqbrk S \in f \sqbrk S$
and:
:$\inf f \sqbrk S \in f \sqbrk S$
Category:Continuous Image of Compact Space is Compact
Continuous Image of Compact Space is Compact/Proof 2

== Theorem ==
== Proof ==
Suppose $\UU$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$.
Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \UU$.
The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \UU$.
Because $T_1$ is compact, it has a finite subcover $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$.
It follows that $\set {U_1, U_2, \ldots, U_r}$ is a finite subcover of $f \sqbrk {T_1}$.
== Sources ==
* : $5$: Compact spaces: $5.5$: Continuous maps on compact spaces: Proposition $5.5.1$
* : $4.14$
Category:Continuous Image of Compact Space is Compact
Continuous Image of Connected Space is Connected

== Theorem ==
Let $T_1$ and $T_2$ be topological spaces.
Let $S_1 \subseteq T_1$ be connected.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then the image $f \sqbrk {S_1}$ is connected.
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $4.22$
Category:Connected Spaces
Category:Continuous Invariants
Category:Continuous Image of Connected Space is Connected
Continuous Image of Connected Space is Connected/Corollary 1

== Corollary to Continuous Image of Connected Space is Connected ==
Connectedness is a topological property.
== Proof ==
Follows directly from Continuous Image of Connected Space is Connected and the definition of topological property.
== Sources ==
* : $6.2$: Connectedness: Corollary $6.2.11$
Category:Connected Spaces
Category:Continuous Image of Connected Space is Connected
Continuous Image of Connected Space is Connected/Corollary 2

== Corollary to Continuous Image of Connected Space is Connected ==
Let $T$ be a connected topological space.
Let $f: T \to \R$ be a continuous real-valued mapping.
Then $f \sqbrk T$ is a real interval.
== Proof ==
From Continuous Image of Connected Space is Connected, the continuous image of the connected space $T$ is connected.
The result follows from Subset of Real Numbers is Interval iff Connected.
== Sources ==
* : $6.2$: Connectedness: Corollary $6.2.13$
Category:Connected Spaces
Category:Continuous Mappings
Category:Continuous Functions
Category:Analysis
Category:Continuous Image of Connected Space is Connected