avewright/proofwiki-math huggingface dataset:
Theorem: Distance from Subset of Real Numbers

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:

Proof:
From the definition of distance:
:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
Thus:
:$\ds \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
Let $\xi = \sup S$.
Then:
:$\forall y \in S: \size {\xi - y} = \xi - y$
So we need to show that no $h > 0$ can be a lower bound for $T = \set {\size {\xi - y}: y \in S}$.
{{AimForCont}} $\exists h > 0: \forall y \in S: \xi - y \ge h$.
But then:
:$\forall y \in S: y \le \xi - h$
and hence $\xi - h$ is an upper bound for $T$ smaller than $\xi = \sup S$.
But this contradicts the definition of supremum, that is the ''smallest'' upper bound.
So there is no such $h > 0$.
Hence by Proof by Contradiction it follows that:
:$\map d {\xi, S} = 0$.
{{Qed}}
Theorem: Distance from Subset of Real Numbers

Let $S$ be a subset of the set of real numbers $\R$.
Let $x \in \R$ be a real number.
Let $\map d {x, S}$ be the distance between $x$ and $S$.
Then:

Proof:
Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.
The result is then seen to be an example of Distance from Subset to Supremum.
{{Qed}}
Theorem: Limit of Sequence to Zero Distance Point

Let $S$ be a non-empty subset of $\R$.
Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.
Then there exists a sequence $\sequence {x_n}$ in $S$ such that $\ds \lim_{n \mathop \to \infty} x_n = \xi$.

Proof:
First it is shown that:
:$\forall n \in \N_{>0}: \exists x_n \in S: \size {\xi - x_n} < \dfrac 1 n$
{{AimForCont}} that:
:$\exists n \in \N_{>0}: \not \exists x \in S: \size {\xi - x} < \dfrac 1 n$
Then $\dfrac 1 n$ is a lower bound of the set $T = \set {\size {\xi - x}: x \in S}$.
This contradicts the assertion that $\map d {\xi, S} = 0$.
We have from Sequence of Powers of Reciprocals is Null Sequence that:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
So as $\size {\xi - x_n} < \dfrac 1 n$ it follows from the Squeeze Theorem for Real Sequences that:
:$\ds \lim_{n \mathop \to \infty} x_n = \xi$
Hence the result.
{{qed}}
Theorem: Image of Closed Real Interval is Bounded

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ is bounded on $\closedint a b$.

Proof:
{{AimForCont}} $f$ is not bounded on $\closedint a b$.
Then from {{Corollary|Limit of Sequence to Zero Distance Point}}, there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
Since $\closedint a b$ is a closed interval, from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_n}$ which converges to some $\xi \in \closedint a b$.
Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
But this contradicts our supposition that there exists a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
The result follows.
{{qed}}
Theorem: Convergent Subsequence in Closed Interval

Let $\closedint a b$ be a closed real interval.
Then every sequence of points of $\closedint a b$ contains a subsequence which converges to a point in $\closedint a b$.

Proof:
Let $\sequence {x_n}$ be a sequence in $\closedint a b$.
Since $\closedint a b$ is bounded in $\R$, it follows that $\sequence {x_n}$ is a bounded sequence.
By the Bolzano-Weierstrass Theorem, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which is convergent.
Suppose $x_{n_r} \to l$ as $n \to \infty$.
Since $a \le x_{n_r} \le b$, from Lower and Upper Bounds for Sequences it follows that $a \le l \le b$.
So $\sequence {x_{n_r} }$ converges to a point in $\closedint a b$.
{{qed}}
Theorem: Max and Min of Function on Closed Real Interval

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ reaches a maximum and a minimum on $\closedint a b$.

Proof:
From Image of Closed Real Interval is Bounded, we have that $f$ is bounded on $\closedint a b$.
Let $d$ be the supremum of $f$ on $\closedint a b$.
Consider a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to d$ as $n \to \infty$.
From {{Corollary|Limit of Sequence to Zero Distance Point}}, this can always be found.
Now $\closedint a b$ is a closed interval
So from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which converges to some $\xi \in \closedint a b$.
Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
So $\map f \xi = d$ and thus the supremum $d$ is indeed a maximum.
A similar argument shows that the infimum is a minimum.
{{qed}}
Theorem: Max and Min of Function on Closed Real Interval

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then $f$ reaches a maximum and a minimum on $\closedint a b$.

Proof:
This is an instance of the Extreme Value Theorem.
$\closedint a b$ is a compact subset of a metric space from Real Number Line is Metric Space.
$\R$ itself is a normed vector space.
{{MissingLinks|Find the reference to $\R$ being a normed vector space.}}
Hence the result.
{{qed}}
Theorem: Continuous Image of Closed Real Interval is Closed Real Interval

Let $f$ be a real function which is continuous on the closed interval $\closedint a b$.
Then the image of $\closedint a b$ under $f$ is also a closed interval.

Proof:
Let $I = \closedint a b$.
Let $J = f \sqbrk I$.
From Image of Real Interval under Continuous Real Function is Real 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.
{{qed}}
Theorem: Retraction Theorem

Let $M$ be a compact manifold with boundary $\partial M$.
Then there is no smooth mapping $f: M \to \partial M$ such that $\partial f: \partial M \to \partial M$ is the identity.

Proof:
{{MissingLinks|one-dimensional, for a start}}
{{AimForCont}} such a smooth mapping exists.
By the Morse-Sard Theorem, there exists a regular value $x \in \partial M$.
By the Preimage Theorem:
:$\map {f^{-1} } x$ is a submanifold of $M$ with boundary.
We have that the codimension of $\map {f^{-1} } x$ in $M$ equals the codimension of $x$ in $\partial M$, that is, $\map \dim M - 1$.
Then $\map {f^{-1} } x$ is one dimensional and compact.
Since $\partial f$ is the identity mapping:
:$\map {\partial f^{-1} } x = \map {f^{-1} } x \cap \partial M = \set x$
This contradicts the Classification of Compact One-Manifolds.
{{qed}}
Category:Algebraic Topology
Category:Named Theorems
ou1wmvvg7a1mlcdvtaln3sfgusjmtpq
Theorem: Classification of Compact One-Manifolds

Every compact connected one-dimensional manifold is diffeomorphic to either a circle or a closed interval.

Proof:
=== Lemma 1 ===
{{:Classification of Compact One-Manifolds/Lemma 1}}{{qed|lemma}}
Let $f$ be a Morse function on a one-manifold $X$.
Let $S$ be the union of the critical points of $f$ and $\partial X$.
As $S$ is finite, $X \setminus S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$.
Theorem: Differentiable Function is Continuous

Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.

Proof:
Let $\map {N_r} 0$ denote the $r$-neighborhood of $0$ in $\C$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {N_r} 0 \setminus \set 0$:
:$(1): \quad \map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$
where $\epsilon: \map {N_r} 0 \setminus \set 0 \to \C$ is a complex function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.
We rewrite the {{RHS}} of $(1)$ to get:
{{begin-eqn}}
{{eqn | l = \map f {a + h}
      | r = \map f a + h \map {f'} a + h \map \epsilon h
}}
{{eqn | ll= \leadsto
      | l = \map f z 
      | r = \map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a}
      | c = substituting $z = a + h$ for $z \in \map {N_r} a$
}}
{{eqn | ll= \leadsto
      | l = \lim_{z \mathop \to a} \map f z
      | r = \lim_{z \mathop \to a} \paren {\map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a} }
      | c = Take the limit on both sides
}}
{{eqn | r = \map f a + \lim_{z \mathop \to a} \paren {z - a} \map {f'} a + \lim_{z \mathop \to a} \paren {\paren {z - a} \map \epsilon {z - a} }
      | c = Combination Theorem for Limits of Complex Functions
}}
{{eqn | r = \map f a
      | c = as $\ds \lim_{z \mathop \to a} \paren {z - a} = 0$, and $\ds \lim_{z \mathop \to a} \map \epsilon {z - a} = 0$
}}
{{end-eqn}}
By definition of continuous complex function, it follows that $f$ is continuous at $a$.
{{qed}}
Theorem: Differentiable Function is Continuous

Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.

Proof:
For each $z \in D$:
{{begin-eqn}}
{{eqn | l = \lim_{w \mathop \to z} \map f w
      | r = \map f z + \lim_{w \mathop \to z} \paren {\map f w - \map f z}
      | c = Sum Rule for Limits of Complex Functions
}}
{{eqn | r = \map f z + \lim_{w \mathop \to z} \paren {\frac {\map f w - \map f z} {w - z} \paren {w - z} }
}}
{{eqn | r = \map f z + \lim_{w \mathop \to z} \frac {\map f w - \map f z} {w - z} \lim_{w \mathop \to z} \paren {w - z}
      | c = Product Rule for Limits of Complex Functions
}}
{{eqn | r = \map f z + \map {f'} z \cdot 0
      | c = {{Defof|Complex-Differentiable Function}}
}}
{{eqn | r = \map f z
}}
{{end-eqn}}
{{qed}}
Theorem: Differentiable Function is Continuous

Let $f$ be a real function defined on an interval $I$.
Let $x_0 \in I$ such that $f$ is differentiable at $x_0$.
Then $f$ is continuous at $x_0$.

Proof:
We have {{hypothesis}} that $\map {f'} {x_0}$ exists.
Let $x, x_0 \in I$ such that $x \ne x_0$.
Then:
{{begin-eqn}}
{{eqn | l = \map f x - \map f {x_0}
      | r = \frac {\map f x - \map f {x_0} } {x - x_0} \cdot \paren {x - x_0}
      | c = 
}}
{{eqn | o = \to
      | r = \map {f'} {x_0} \cdot 0
      | c = as $x \to x_0$
}}
{{end-eqn}}
Thus:
:$\map f x \to \map f {x_0}$ as $x \to x_0$
or in other words:
:$\ds \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$
The result follows by definition of continuous.
{{qed}}
Theorem: Preimage Theorem

Let $y$ be a regular value of a smooth submersion $f: X \to Y$.
Then the preimage $\map {f^{-1} } y$ is a smooth submanifold of $X$, with $\dim \map {f^{-1} } y = \dim X - \dim Y$.

Proof:
Let $k, l$ be natural numbers with $k \ge l$.
By the Local Submersion Theorem, there exists coordinates in some open neighborhoods of $x, y$ such that $\map f {x_1, x_2, \ldots, x_k} = \tuple {x_1, \ldots, x_l}$ and $y$ corresponds to $\tuple {0, \ldots, 0}$.
Let $V$ be that neighborhood of $x$.
Then $\map {f^{-1} } y \cap V$ is the set of points where $x_1 = 0, \ldots, x_l = 0$.
The functions $x_{l + 1}, \ldots, x_k$ therefore form a coordinate system on the set $\map {f^{-1} } y \cap V$, which is a relatively open subset of $\map {f^{-1} } y$.
Together these functions then form a diffeomorphism to a Euclidean space.
We also have, by the regular value properties of $y$, a surjection of tangent spaces from $x$ to $y$.
This ensures smoothness of the solution set $\map {f^{-1} } y$.
{{qed}}
Theorem: Forward-Backward Induction

Let $P$ be a propositional function on the natural numbers $\N$.
Suppose that:
:$(1): \quad \forall n \in \N: \map P {2^n}$ holds.
:$(2): \quad \map P n \implies \map P {n - 1}$.
Then $\map P n$ holds for all $\forall n \in \N$.
The proof technique based on this result is called '''forward-backward induction'''.

Proof:
{{AimForCont}} $\exists k \in \N$ such that $\map P k$ is false.
From Power of Real Number greater than One is Unbounded Above:
:$\set {2^n: n \in \N}$ is unbounded above.
Therefore we can find:
:$M = 2^N > k$
Now let us create the set:
:$S = \set {n \in \N: n < M, \map P n \text { is false} }$
As $k < M$ and $\map P k$ is false:
:$S \ne \O$
As $\forall x \in S: x < M$ it follows that $S$ is bounded above.
So from Set of Integers Bounded Above by Integer has Greatest Element, $S$ has a greatest element, which we call $m$.
We have that $\map P n$ holds for $m < n \le M$.
Hence $\map P {m + 1}$ holds.
But $\map P {m + 1} \implies \map P m$.
This contradicts our assertion that $\map P n$ is false.
Hence by Proof by Contradiction there can be no such $k \in \N$ such that $\map P k$ is false.
Hence the result.
{{qed}}
Theorem: Inequalities Concerning Roots

Let $\closedint X Y$ be a closed real interval such that $0 < X \le Y$.
Let $x, y \in \closedint X Y$.
Then:
:$\forall n \in \N_{> 0}: X Y^{1/n} \size {x - y} \le n X Y \size {x^{1/n} - y^{1/n} } \le Y X^{1/n} \size {x - y}$

Proof:
From Difference of Two Powers:
:$\ds a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
Let $a > b > 0$.
Then:
:$\dfrac {a^n - b^n} {a - b} = a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1}$
Also, note that:
:$b^{n - 1} = b^{n - j - 1} b^j \le a^{n - j - 1} b^j \le a^{n - j -1} a^j \le a^{n - 1}$
Hence if $a > b > 0$:
:$n b^{n - 1} \le \dfrac {a^n - b^n} {a - b} \le n a^{n - 1}$
Taking reciprocals:
:$\dfrac 1 {n b^{n - 1} } \ge \dfrac {a - b} {a^n - b^n} \ge \dfrac 1 {n a^{n - 1} }$
Thus:
:$\dfrac {a^n - b^n} {n a^{n - 1} } \le a - b \le \dfrac {a^n - b^n} {n b^{n - 1} } \implies \dfrac {a \paren {a^n - b^n} } {a^n} \le n \paren {a - b} \le \dfrac {b \paren {a^n - b^n} } {b^n}$
Now suppose $x > y$.
Then:
:$x - y = \size {x - y}$
and:
:$x^{1/n} - y^{1/n} = \size {x^{1/n} - y^{1/n} }$
Then put $a = x^{1/n}$ and $b = y^{1/n}$ in the above:
:$\dfrac {x^{1/n} \size {x - y} } x \le n \size {x^{1/n} - y^{1/n} } \le \dfrac {y^{1/n} \size {x - y} } y$
Multiplying through by $x y$:
:$x^{1/n} y \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le x y^{1/n} \size {x - y}$
Similarly, if $y > x$:
:$y^{1/n} x \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le y x^{1/n} \size {x - y}$
The result follows after some algebra.
{{qed}}
Theorem: Continuity of Root Function

Let $n \in \N_{>0}$ be a non-zero natural number.
Let $f: \hointr 0 \infty \to \R$ be the real function defined by $\map f x = x^{1/n}$.
Then $f$ is continuous at each $\xi > 0$ and continuous on the right at $\xi = 0$.

Proof:
First suppose that $\xi > 0$.
Let $X, Y \in \R$ such that $0 < X < \xi < Y$.
Let $x \in \R$ such that $X < x < Y$.
From Inequalities Concerning Roots:
:$X Y^{1/n} \ \size {x - \xi} \le n X Y \ \size {x^{1/n} - \xi^{1/n} } \le Y X^{1/n} \ \size {x - \xi}$
Thus:
:$\dfrac 1 {n Y} Y^{1/n} \ \size {x - \xi} \le \size {x^{1/n} - \xi^{1/n} } \le \dfrac 1 {n X} X^{1/n} \ \size {x - \xi}$
The result follows by applying the Squeeze Theorem.
Now we need to show that $\map f x \to 0$ as $x \to 0^+$.
We need to show that:
:$\forall \epsilon > 0: \exists \delta > 0: x^{1/n} = \size {x^{1/n} - 0} < \epsilon$
provided $0 < x < \delta$.
Clearly, for any given $\epsilon$, we can choose $\delta = \epsilon^n$.
Hence the result.
{{qed}}
Theorem: Weak Whitney Immersion Theorem

Every $k$-dimensional manifold $X$ admits a one-to-one immersion in $\R^{2 k + 1}$.
{{MissingLinks|$k$-dimensional}}

Proof:
Let $M > 2 k + 1$ be a natural number such that $f: X \to \R^M$ is an injective immersion.
Define a map $h: X \times X \times \R \to \R^M$ by:
:$\map h {x, y, t} = \map t {\map f x - \map f y}$
Define a map $g: \map T X \to \R^M$ by:
:$\map g {x, v} = \map {\d f_x} v$
where $\map T X$ is the tangent bundle of $X$.
Since $M > 2 k + 1$, the Morse-Sard Theorem implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.
Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.  
Suppose:
:$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$
for some $x, y$.
Then:
:$\map f x - \map f y = t a$
for some scalar $t$.
{{AimForCont}} $x \ne y$.
Then as $f$ is injective:
:$t \ne 0$
But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$.
By Proof by Contradiction:
:$x = y$
and so $\pi \circ f$ is injective.
Let $v$ be a nonzero vector in $\map {T_x} X$ (the tangent space of $X$ at $x$) for which:
:$\map {\d \paren {\pi \circ f}_x} v = 0$
Since $\pi$ is linear:
:$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$
Thus:
:$\map {\pi \circ \d f_x} v = 0$
so:
:$\map {\d f_x} v = t a$
for some scalar $t$.
Because $f$ is an immersion, $t \ne 0$.
Hence:
:$\map g {x, 1/t} = a$
contradicting the choice of $a$.
Hence $\pi \circ f: X \to H$ is an immersion. 
$H$ is obviously isomorphic to $\R^{M-1}$.
Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.
Theorem: Fundamental Theorem of Algebra

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
That is, the field of complex numbers is algebraically closed.

Proof:
Let $\map p z$ be a polynomial in $\C$:
:$\map p z = z^m + a_1 z^{m-1} + \cdots + a_m$
where not all of $a_1, \ldots, a_m$ are zero.
Define a homotopy:
:$\map {p_t} z = t \map p z + \left({1-t}\right) z^m$
Then:
:$\dfrac {\map {p_t} z} {z^m} = 1 + t \paren {a_1 \dfrac 1 z + \cdots + a_m \dfrac 1 {z^m} }$
The terms in the parenthesis go to $0$ as $z \to \infty$.
{{explain}}
Therefore, there is an $r \in \R_{>0}$ such that:
:$\forall z \in \C: \size z = r: \forall t \in \closedint 0 1: \map {p_t} z \ne 0$
Hence the homotopy:
:$\dfrac {p_t} {\size {p_t} }: S \to \Bbb S^1$
is well-defined for all $t$.
This shows that for any complex polynomial $\map p z$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac {\map p z} {\size {\map p z}}$ and $\dfrac {z^m} {\size {z^m} }$ are freely homotopic maps $S \to \Bbb S^1$. 
Hence $\dfrac {\map p z} {\size {\map p z} }$ must have the same degree of $\paren {z / r}^m$, which is $m$.  
When $m > 0$, that is $p$ is non-constant, this result and the Extendability Theorem for Intersection Numbers imply $\dfrac {\map p z} {\size {\map p z} }$ does not extend to the disk $\Int S$, implying $\map p z = 0$ for some $z \in \Int S$.
{{explain|$\Int S$}}
{{qed}}
Theorem: Fundamental Theorem of Algebra

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
That is, the field of complex numbers is algebraically closed.

Proof:
Let $\map P z = a_n z^n + \dots + a_1 z + a_0, \ a_n \ne 0$.
{{AimForCont}} that $\map P z$ is not zero for any $z \in \C$.
It follows that $1 / \map P z$ must be entire; and is also bounded in the complex plane.
In order to see that it is indeed bounded, we recall that $\exists R \in \R_{>0}$ such that:
   $\cmod {\dfrac 1 {\map P z} } < \dfrac 2 {\cmod {a_n} R^n}, \text{whenever} \ \cmod z > R$.
Hence, $1 / \map P z$ is bounded in the region outside the disk $\cmod z \le R$.
However, $1 / \map P z$ is continuous on that closed disk, and thus it is bounded there as well. 
Furthermore, we observe that $1 / \map P x$ must be bounded in the whole plane.
Through Liouville's Theorem, $1 / \map P x$, and thus $\map P x$, is constant. 
This is a contradiction.
{{qed}}
Theorem: Fundamental Theorem of Algebra

Every non-constant polynomial with coefficients in $\C$ has a root in $\C$.
That is, the field of complex numbers is algebraically closed.

Proof:
Let $p: \C \to \C$ be a complex, non-constant polynomial.
{{AimForCont}} that $\map p z \ne 0$ for all $z \in \C$.
Now consider the closed contour integral:
:$\ds \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$
where $\gamma_R$ is a circle with radius $R$ around the origin.
By Derivative of Complex Polynomial, the polynomial $z \cdot \map p z$ is holomorphic.
Since $\map p z$ is assumed to have no zeros, the only zero of $z \cdot \map p z$ is $0 \in \C$.
Therefore by Reciprocal of Holomorphic Function $\dfrac 1 {z \cdot \map p z}$ is holomorphic in $\C \setminus \set 0$.
Hence the Cauchy-Goursat Theorem implies that the value of this integral is independent of $R > 0$.
On the one hand, one can calculate the value of this integral in the limit $R \to 0$ (or use Cauchy's Residue Theorem), using the parameterization $z = R e^{i \phi}$ of $\gamma_R$:
{{begin-eqn}}
{{eqn | l = \lim \limits_{R \mathop \to 0} \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z
      | r = \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac 1 {R e^{i \phi} {\map p {R e^{i \phi} } } } \, i R e^{i \phi} \rd \phi
      | c = {{Defof|Complex Contour Integral}}
}}
{{eqn | r = \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac i {\map p {R e^{i \phi} } } \rd \phi
}}
{{eqn | r = \int \limits_0^{2 \pi} \lim \limits_{R \mathop \to 0} \frac i {\map p {R e^{i \phi} } } \rd \phi
      | c = Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions
}}
{{eqn | r = \int \limits_0^{2 \pi} \frac i {\map p 0} \rd \phi
      | c = Real Polynomial Function is Continuous
}}
{{eqn | r = \frac {2 \pi i} {\map p 0}
}}
{{end-eqn}}
which is non-zero.
On the other hand, we have the following upper bound for the absolute value of the integral:
{{begin-eqn}}
{{eqn | l = \size {\oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z}
      | o = \le
      | r = 2 \pi R \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } }
      | c = Estimation Lemma for Contour Integrals
}}
{{eqn | r = 2 \pi \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {\map p z} } }
}}
{{end-eqn}}
But this goes to zero for $R \to \infty$.
We have arrived at a contradiction.
Hence by Proof by Contradiction the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.
{{qed}}