nyuuzyou/wikis huggingface dataset:
Space of Piecewise Linear Functions on Closed Interval is Dense in Space of Continuous Functions on Closed Interval

== Theorem ==
Let $I = \closedint a b$. 
Let $\map \CC I$ be the set of continuous functions on $I$. 
Let $\map {\mathrm {PL} } I$ be the set of piecewise linear functions on $I$. 
Let $d$ be the metric induced by the supremum norm.
Then $\map {\mathrm {PL} } I$ is dense in $\struct {\map \CC I, d}$.
== Proof ==
Let $f \in \map \CC I$. 
Let $\epsilon \in \R_{>0}$ be a real number.
From Open Ball Characterization of Denseness:
:it suffices to find a $p \in \map {\mathrm {PL} } I$ such that $p$ is contained in the open ball $\map {B_\epsilon} f$.
From Continuous Function on Closed Real Interval is Uniformly Continuous:
:$f$ is uniformly continuous on $I$.
That is: 
:there exists a $\delta > 0$ such that for all $x, y \in I$ with $\size {x - y} < \delta$ we have $\size {\map f x - \map f y} < \epsilon/3$.
Let:
:$P = \{a_0 = a, a_1, a_2, \ldots, a_n = b\}$
be a finite subdivision of $I$, with: 
:$\size {a_{i + 1} - a_i} < \delta$
for each $i$.
Let $p \in \map {\mathrm {PL} } I$ be such that: 
:$\map p {a_i} = \map f {a_i}$
for each $i$, with $p$ continuous. 
We can explicitly construct such a $p$ by connecting $\tuple {a_i, \map f {a_i} }$ to $\tuple {a_{i + 1}, \map f {a_{i + 1} } }$ with a straight line segment for each $i$.
Fix $x \in I$.
Note that there exists precisely one $i$ such that $a_i \le x \le a_{i + 1}$, fix this $i$.
We then have: 
since $\size {a_{i + 1} - a_i} < \delta$.
Since $\size {x - a_i} < \size {a_{i + 1} - a_i} < \delta$, we also have: 
:$\size {\map f x - \map f {a_i} } < \epsilon/3$
So:
Note that $x \in I$ was arbitrary, so:
so $p \in \map {B_\epsilon} f$.
Category:Functional Analysis
Space of Real-Valued Measurable Functions Identified by A.E. Equality is Vector Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map \MM {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim_\mu$ be the almost-everywhere equality relation on $\map \MM {X, \Sigma, \R}$ with respect to $\mu$.
Let $+$ denote pointwise addition on $\map \MM {X, \Sigma, \R} / \sim_\mu$.
Let $\cdot$ be pointwise scalar multiplication on $\map \MM {X, \Sigma, \R} / \sim_\mu$.
Then: 
:$\struct {\map \MM {X, \Sigma, \R} / \sim_\mu, +, \cdot}_\R$
forms a vector space.
== Proof ==
We verify each of the vector space axioms.
=== ===
This is shown in Pointwise Addition on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined.
=== ===
Let $E_1, E_2 \in \map \MM {X, \Sigma, \R} / \sim_\mu$.
Write $E_1 = \eqclass f \sim$ and $E_2 = \eqclass g \sim$. 
Then: 
=== ===
Let $E_1, E_2, E_3 \in \map \MM {X, \Sigma, \R} / \sim_\mu$.
Write $E_1 = \eqclass {f_1} \sim$, $E_2 = \eqclass {f_2} \sim$ and $E_3 = \eqclass {f_3} \sim$. 
Then we have: 
=== ===
Set: 
:$\mathbf 0 = \eqclass 0 \sim$ 
Let $E \in \map \MM {X, \Sigma, \R} / \sim_\mu$. 
Then, write $E = \eqclass f \sim$. 
Then we have: 
Combining this with $(\text V 2)$, we get: 
:$\mathbf 0 + E = E + \mathbf 0 = E$
=== ===
Let $E \in \map \MM {X, \Sigma, \R} / \sim_\mu$. 
Then, write $E = \eqclass f \sim$. 
Let: 
:$-E = \paren {-1} \cdot E$
Then we have: 
=== ===
Let $E \in \map \MM {X, \Sigma, \R} / \sim_\mu$ and $\lambda, \mu \in \R$.
Then, write $E = \eqclass f \sim$. 
Then, we have: 
=== ===
Let $E_1, E_2 \in \map \MM {X, \Sigma, \R} / \sim_\mu$ and $\lambda \in \R$.
Write $E_1 = \eqclass f \sim$ and $E_2 = \eqclass g \sim$. 
Then: 
=== ===
Let $E \in \map \MM {X, \Sigma, \R} / \sim_\mu$ and $\lambda, \mu \in \R$.
Write $E = \sim f \sim$. 
Then we have: 
=== ===
Let $E \in \map \MM {X, \Sigma, \R} / \sim_\mu$.
Write $E = \sim f \sim$.
Then we have: 
Category:Vector Spaces
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
Space of Simple P-Integrable Functions is Everywhere Dense in Lebesgue Space

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
Let $\map {\LL^p} \mu$ be Lebesgue $p$-space for $\mu$.
Let $\map \EE \Sigma \cap \map {\LL^p} \mu$ be the space of $\Sigma$-simple, $p$-integrable functions.
Then $\map \EE \Sigma \cap \map {\LL^p} \mu$ is everywhere dense in $\map {\LL^p} \mu$ with respect to the seminorm topology generated by the singleton $\set {\norm {\; \cdot \;}_p }$.
That is, for all $f \in \map {\LL^p} \mu$ and $\epsilon > 0$ there exists a $g \in \map \EE \Sigma \cap \map {\LL^p} \mu$ such that:
:$\norm {f - g}_p < \epsilon$
== Proof ==
For $n \in \N$, we define $D_n : \R_{\ge 0} \to \R_{\ge 0}$ by:
$\quad \map {\Delta _n} y := \begin{cases}
\dfrac k {2^n} & : y \in \hointr {\dfrac k {2^n} } {\dfrac {k + 1} {2^n} }, k = 0, 1, \ldots, 2^{2 n} - 1 \\
0 & : y \ge 2^n
\end{cases}$
Clearly, for all $n \in \N$:
:$(1): \quad \forall y \in \R_{\ge 0} : 0 \le \map {\Delta_n} y \le y$
We also have:
:$(2): \quad \ds \forall y \in \R_{\ge 0} : \lim_{n \mathop \to \infty} \map {\Delta_n} y = y$
since:
:$\forall y \in \hointr 0 {2^n} : 0 \le y - \map {\Delta_n} y \le \dfrac 1 {2^n}$
Let:
:$\map {f_n} x := \map \sgn {\map f x} \map {\Delta_n} {\size {\map f x} }$
where:
:$\map \sgn \cdot$ denotes the signum function
:$\size {\, \cdot \,}$ denotes the absolute value.
By Measurable Function is Simple Function iff Finite Image Set:
:$f_n \in \map \EE \Sigma$
By $(2)$, for all $x \in X$:
:$\ds \lim_{n \mathop \to \infty} \map {f_n} x = \map f x$
In view of $(1)$, we have for all $x \in X$:
:$\size {\map {f_n} x} \le \size {\map f x}$
In particular, $f_n \in \map {\LL^p} \mu$.
Moreover, for all $x \in X$:
:$\size {\map {f_n} x - \map f x} \le \size {\map f x} + \size {\map {f_n} x} \le 2 \size {\map f x}$
Thus, by Lebesgue's Dominated Convergence Theorem:
:$\ds \lim_{n \mathop \to \infty} \norm {f_n - f}_p = \paren {\int \lim_{n \mathop \to \infty} \size {f_n - f}^p \rd \mu }^{1/p} = 0$
== Sources ==
* : $12.11$
Category:Lebesgue Spaces
Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval

== Theorem ==
Let $I = \closedint a b$. 
Let $\map \CC I$ be the set of continuous functions on $I$. 
Let $\map \DD I$ be the set of continuous functions on $I$ that are differentiable at a point.
Let $d$ be the metric induced by the supremum norm.
Then $\map \DD I$ is meager in $\struct {\map \CC I, d}$.
=== Corollary ===
== Proof ==
Let: 
:$\ds A_{n, m} = \set {f \in \map \CC I: \exists x \in I: \forall t \in I: 0 < \size {t - x} < \frac 1 m \implies \size {\frac {\map f t - \map f x} {t - x} } \le n}$
and:
:$\ds A = \bigcup_{\tuple {n, m} \mathop \in \N^2} A_{n, m}$
=== Lemma 1 ===
=== Lemma 2 ===
From Lemma 2: 
:$A$ is the countable union of nowhere dense sets.
So, by definition of a meager space: 
:$A$ is meager in $\struct {\map \CC I, d}$.
By Subset of Meager Set is Meager Set and Lemma 1: 
:$\map \DD I$ is meager in $\struct {\map \CC I, d}$.
Category:Functional Analysis
Category:Meager Spaces
Category:Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval
Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Corollary

== Theorem ==
Let $I = \closedint a b$. 
Let $\map \CC I$ be the set of continuous functions on $I$. 
Then:
:there exists a function $f \in \map \CC I$ that is not differentiable anywhere.
== Proof ==
Let $\map \DD I$ be the set of continuous functions on $I$ that are differentiable at a point.
Let $d$ be the metric induced by the supremum norm.
By Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Banach Space:
:$\struct {\map \CC I, d}$ is a complete metric space.
By Baire Space is Non-Meager:
:$\map \CC I$ is non-meager in $\struct {\map \CC I, d}$.
By Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval:
:$\map \DD I$ is meager in $\struct {\map \CC I, d}$.
So:
:$\map \DD I \ne \map \CC I$.
That is, there exists a continuous function that is not differentiable anywhere.
== Examples ==
An explicit example is given in Weierstrass's Theorem.
Category:Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval
Category:Functional Analysis
Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 1

== Theorem ==
Let $I = \closedint a b$.
Let $\map \CC I$ be the set of continuous functions on $I$. 
Let $\map \DD I$ be the set of continuous functions on $I$ that are differentiable at a point.
Let: 
:$\ds A_{n, m} = \set {f \in \map \CC I: \text {there exists } x \in I \text { such that } \size {\frac {\map f t - \map f x} {t - x} } \le n \text { for all } t \text { with } 0 < \size {t - x} < \frac 1 m}$
and:
:$\ds A = \bigcup_{\tuple {n, m} \in \N^2} A_{n, m}$
Then:
:$\map \DD I \subseteq A$
== Proof ==
Let $f \in \map \DD I$. 
Then, $f$ is differentiable at some $x \in I$.
Let: 
:$n = \floor {\size {\map {f'} x} } + 1$
where $\floor \cdot$ is the floor function.
Then:
:$\size {\map {f'} x} < n$
From the definition of the derivative, there exists $\delta > 0$ such that for all $t$ with $0 < \size {t - x} < \delta$, we have: 
:$\ds \size {\frac {\map f t - \map f x} {t - x} - \map {f'} x} < 1 - \fractpart {\size {\map {f'} x} }$
From the Reverse Triangle Inequality, we then have: 
:$\ds \size {\size {\frac {\map f t - \map f x} {t - x} } - \size {\map {f'} x} } < 1 - \fractpart {\size {\map {f'} x} }$
and so:
for all $t$ with $0 < \size {t - x} < \delta$.
Pick $m \in \N$ such that $\frac 1 m < \delta$.
We then have:
:$\ds \size {\frac {\map f t - \map f x} {t - x} } \le n$
for $t$ with $0 < \size {t - x} < \frac 1 m$.
That is:
:$f \in A_{n, \, m} \subseteq A$
so:
:$\map \DD I \subseteq A$
as required.
Category:Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval
Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval/Lemma 2/Lemma 2.1

== Theorem ==
Let $I = \closedint a b$.
Let $\map \CC I$ be the set of continuous functions on $I$. 
Let $d$ be the metric induced by the supremum norm.
Let: 
:$\ds A_{n, m} = \set {f \in \map \CC I: \exists x \in I: \forall t \in \R: 0 < \size {t - x} < \frac 1 m \implies \size {\frac {\map f t - \map f x} {t - x} \le n} }$
Then:
:for each $\tuple {n, m} \in \N^2$, $A_{n, m}$ is closed in $\tuple {\map \CC I, d}$. 
== Proof ==
Fix $\tuple {n, m} \in \N^2$. 
From Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Banach Space:
:$\tuple {\map \CC I, d}$ is complete.
Hence, from Subspace of Complete Metric Space is Closed iff Complete:
:$A_{n, m}$ is closed $\tuple {A_{n, m}, d}$ is complete.
Let $\sequence {f_i}_{i \mathop \in \N}$ be a Cauchy sequence in $\tuple {A_{n, m}, d}$.
Since $\tuple {\map \CC I, d}$ is complete, $\sequence {f_i}_{i \mathop \in \N}$ converges some $f \in \map \CC I$.
We aim to show that $f \in A_{n, m}$.
Since $f_i \in A_{n, m}$ for each $i \in \N$, there exists $x_i \in I$ such that: 
:$\ds \size {\frac {\map {f_i} t - \map {f_i} {x_i} } {t - x_i} } \le n$ for each $t$ with $0 < \size {t - x_i} < \dfrac 1 m$.
Note that since $I$ is bounded, $\sequence {x_i}_{i \mathop \in \N}$ is bounded.
Therefore, by the Bolzano-Weierstrass Theorem:
:there exists a convergent subsequence of $\sequence {x_i}_{i \mathop \in \N}$, $\sequence {x_{i_k} }_{k \mathop \in \N}$.
Let: 
:$\ds x = \lim_{k \mathop \to \infty} x_{i_k}$
Note that we also have: 
:$\ds f = \lim_{k \mathop \to \infty} f_{i_k}$
From Subset of Metric Space contains Limits of Sequences iff Closed, since $I$ is closed, $x \in I$.
From Necessary Condition for Uniform Convergence:
:the sequence $\sequence {\map {f_{i_k} } {x_{i_k} } }_{k \mathop \in \N}$ converges to $\map f x$.
We therefore have: 
for all $t$ with $0 < \size {t - x} < \dfrac 1 m$. 
That is, $f \in A_{n, m}$.
Category:Space of Somewhere Differentiable Continuous Functions on Closed Interval is Meager in Space of Continuous Functions on Closed Interval
Space of Square Summable Mappings is Hilbert Space

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $I$ be a set.
Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.
Let $\innerprod \cdot \cdot: \map {\ell^2} I \times \map {\ell^2} I \to \GF$ be the inner product on $\map {\ell^2} I$.
Then $\map {\ell^2} I$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.
== Proof ==
By Space of Square Summable Mappings is $L^2$ Space, $\map {\ell^2} I$ is equal to $\map {L^2} {I, \powerset I, \mu}$.
The result follows by $L^2$ Space forms Hilbert Space.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.7$
Category:Examples of Hilbert Spaces
Category:Space of Square Summable Mappings
Space of Square Summable Mappings is L-2 Space

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $I$ be a set.
Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.
Then $\map {\ell^2} I$ is equal to the $L^2$ space $\map {L^2} {I, \powerset I, \mu}$, where $\mu$ is the counting measure on $I$.
== Proof ==
First, let us unpack the definition of the Lebesgue space $\map {\LL^2} {I, \powerset I, \mu}$:
:$\ds \map {\LL^2} {I, \powerset I, \mu} = \set{ f: I \to \GF: f \in \map \MM { \powerset I }, \int \size f^2 \rd \mu < \infty}$
By Function Measurable with respect to Power Set, all mappings $f: I \to \GF$ are measurable.
By Integral over Counting Measure is Sum of Values, $\ds \int \size f^2 \rd \mu = \sum_{i \mathop \in I} \cmod{ \map f i }^2$.
Hence:
:$\ds \map {\LL^2} {I, \powerset I, \mu} = \set{ f: I \to \GF: \sum_{i \mathop \in I} \cmod{ \map f i }^2 < \infty}$
which is exactly the definition of $\map {\ell^2} I$.
Next, the definition of $\map {L^2} {I, \powerset I, \mu}$ is:
:$\map {L^2} \mu = \map {\LL^2} {\mu} / \sim_\mu$
where $\sim_\mu$ is the $\mu$-almost-everywhere equality relation.
Now let $f, g \in \map {\ell^2} I$ such that $\map \mu { \set{ f \ne g } } = 0$.
Then since $\mu$ is the counting measure, it follows that $\set{ f \ne g } = \varnothing$.
That is, $f = g$.
Therefore, $\sim_\mu$ is the trivial equivalence and $\map {\ell^2} I = \map {L^2} {I, \powerset I, \mu}$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples: Example $1.7$
Category:Lp Spaces
Category:Space of Square Summable Mappings
Space of Square Summable Mappings is Vector Space

== Theorem ==
Let $\GF$ be a subfield of $\C$.
Let $I$ be a set.
Let $\map {\ell^2} I$ be the space of square summable mappings over $I$.
Then $\map {\ell^2} I$ is a vector space.
== Proof ==
By definition, $\map {\ell^2} I$ is a subset of the vector space $\GF^I$ of all mappings $f: I \to \GF$.
Let us apply the One-Step Vector Subspace Test.
Thus, let $f, g \in \map {\ell^2} I$ and $\lambda \in \GF$.
Then we must show that $f + \lambda g: I \to \GF$ is square summable.
First, note that:
:$\set{ i \in I: \map f i + \lambda \map g i \ne 0 } \subseteq \set{ i \in I: \map f i \ne 0 } \cup \set{ i \in I: \map g i \ne 0 }$
By Finite Union of Countable Sets is Countable and Subset of Countable Set is Countable, it follows that:
:$\set{ i \in I: \map f i + \lambda \map g i \ne 0 }$
is countable.
Next:
so that $f + \lambda g$ is square summable.
Hence the result by the One-Step Vector Subspace Test.
Category:Examples of Vector Spaces
Category:Space of Square Summable Mappings
Space of Zero-Limit Sequences with Supremum Norm forms Banach Space

== Theorem ==
Let $c_0$ be the space of zero-limit sequences.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm.
Then $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.
== Proof ==
Let $\sequence {a_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.
Let $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.
By Space of Zero-Limit Sequences with Supremum Norm forms Normed Vector Space, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a subspace of $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.
Hence, $\sequence {a_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.
By Space of Bounded Sequences with Supremum Norm forms Banach Space, $\sequence {a_n}_{n \mathop \in \N}$ converges to $a \in \ell^\infty$.
Denote $a_n = \sequence {a_n^{\paren m}}_{m \mathop \in \N}$ and $a = \sequence {a^{\paren m}}_{m \mathop \in \N}$.
By definition of convergent sequences:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {a_n - a}_\infty < \epsilon$
Then:
Since $a_n \in c_0$:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists M \in \R_{\mathop > 0} : \forall m \in \N : m > M \implies \size {a_n^{\paren m}} < \epsilon$
For all $m > M$ we also have that:
In other words:
:$\forall \epsilon' \in \R_{>0} : \exists M' \in \R_{>0} : \forall m \in \N : m > M' \implies \size {a^{\paren m} } < \epsilon'$
where $\epsilon' = 2\epsilon$ and $M' = M$.
By definition of zero-limit sequences, $a \in c_0$.
Therefore, in $\struct {c_0, \norm {\, \cdot \,}_\infty}$ a Cauchy sequence is also convergent in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.
By definition, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Banach Spaces
Space such that Intersection of Open Sets containing Point is Singleton may not be Hausdorff

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be arbitrary.
Let $T$ be such that the intersection of all open sets containing $x$ is $\set x$:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$
Then it is not necessarily the case that $T$ is a Hausdorff space.
== Proof ==
Let $T$ be the finite complement topology on the real numbers $\R$, for example.
The open sets of $T$ are subsets of $\R$ of the form $U$ such that $\R \setminus U$ is finite, together with $\O$.
Let $x \in \R$ be arbitrary.
Let $K = \ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} }$, that is, the intersection of all open sets of $T$ containing $x$.
Let $y \in \R$ such that $y \ne x$.
Note that the set $\set y$ is finite.
Thus $\R \setminus \set y$ is an open set of $T$ which contains $x$ but not $y$.
Hence $y \notin K$.
As $y$ is arbitrary, it follows that the only element of $\R$ which is in $K$ is $x$ itself.
That is:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$
But from Finite Complement Space is not Hausdorff, $T$ is not a Hausdorff space.
== Also see ==
* Intersection of Open Sets of Hausdorff Space containing Point is Singleton: the motivation behind this result
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 4 \ \text {(b)}$
Category:Hausdorff Spaces
Space with Open Point is Non-Meager

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be an open point.
Then $T$ is a non-meager space.
== Proof ==
Let $x \in S$ be an open point of $T$.
That is:
:$\set x \in \tau$
Recall that:
:a topological space is non-meager if it is not meager
and:
:a topological space is meager it is a countable union of subsets of $S$ which are nowhere dense in $S$.
 that $T$ is meager.
Let:
:$\ds T = \bigcup \SS$
where $\SS$ is a countable set of subsets of $S$ which are nowhere dense in $S$.
Then:
:$\exists H \in \S S: x \in H$
and so:
:$\set x \subseteq H$
We have that $H$ is nowhere dense in $T$.
By definition, its closure $H^-$ contains no open set of $T$ which is non-empty.
But from Set is Subset of its Topological Closure we have that:
:$H \subseteq H^-$
So by Subset Relation is Transitive:
:$\set x \subseteq H^-$
So $H$ is not nowhere dense.
Therefore $T$ cannot be a countable union of subsets of $S$ which are nowhere dense in $S$.
That is, $T$ is not meager.
Hence the result by definition of non-meager.
== Sources ==
* : Problems: Section $1: \ 4$
Category:Non-Meager Spaces
Spaces of Continuous Functions on Homeomorphic Compact Hausdorff Spaces are Isometrically *-Algebra Isomorphic

== Theorem ==
Let $K, L$ be homeomorphic compact Hausdorff spaces.
Let $\map \CC K$ and $\map \CC L$ be the spaces of continuous functions on $K$ and $L$ respectively.
Then $\map \CC K$ and $\map \CC L$ are isometrically $\ast$-algebra isomorphic.
== Proof ==
Let $f : K \to L$ be a homeomorphism.
Let $\phi \in \map \CC L$.
From Composite of Continuous Mappings is Continuous, $\phi \circ f \in \map \CC K$.
Hence we can define $T : \map \CC L \to \map \CC K$ by: 
:$T \phi = \phi \circ f$
for each $\phi \in \map \CC L$.
Let $\phi, \psi \in \map \CC L$, $\lambda \in \C$ and $x \in K$. 
Then we have:
hence:
:$T \phi + \lambda T \psi = \map T {\phi + \lambda \psi}$
So $T$ is linear. 
We also have:
Hence we obtain:
:$\paren {T \phi} \paren {T \psi} = \map T {\phi \psi}$
Hence $T$ is an algebra homomorphism.
Additionally we have:
Hence:
:$\overline {T \phi} = T \overline \phi$
So $T$ is a $\ast$-algebra homomorphism.
We now need to establish that $T$ is an algebra isomorphism.
Note that since $f$ is a bijection, if $\phi, \psi \in \map \CC L$ are such that:
:$\phi \circ f = \psi \circ f$
we have that:
:$\phi = \psi$ 
Hence $T$ is an injection.
Let $u \in \map \CC K$.
Further, since $f^{-1}$ is continuous, we obtain that $u \circ f^{-1} \in \map \CC L$ from Composite of Continuous Mappings is Continuous.
Hence we obtain:
:$\map T {u \circ f^{-1} } = \paren {u \circ f^{-1} } \circ f = u$ 
So $T$ is surjective.
So $T$ is indeed an algebra isomorphism.
We finally show that $T$ is an isometry. 
By the definition of the supremum norm, we wish to show that:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
We note that since $\map f x \in L$ for each $x \in K$:
:$\ds \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} } \le \sup_{y \mathop \in L} \cmod {\map \phi y}$
From the extreme value theorem for normed vector spaces, since $L$ is compact, there exists $y_\ast \in L$ such that:
:$\ds \cmod {\map \phi {y_\ast} } = \sup_{y \mathop \in L} \cmod {\map \phi y}$
Since $f$ is a bijection, there exists $x_\ast \in K$ such that $\map f {x_\ast} = y_\ast$ and:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \cmod {\map \phi {\map f {x_\ast} } } \le \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
So we obtain:
:$\ds \sup_{y \mathop \in L} \cmod {\map \phi y} = \sup_{x \mathop \in K} \cmod {\map \phi {\map f x} }$
Hence $T$ is an isometric $\ast$-algebra isomorphism.
Category:*-Algebras
Category:Space of Continuous Functions on Compact Hausdorff Space