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If $f$ is holomorphic on the open ball of radius $r$ centered at $0$ and $f(0) = 0$, then there exists a holomorphic function $h$ on the open ball of radius $r$ centered at $0$ such that for all $z$ with $|z| < r$, we have $f(z) = zh(z)$ and $f'(0) = h(0)$.
If $f$ is a holomorphic function on the unit disk such that $f(0) = 0$ and $|f(z)| < 1$ for all $z$ in the disk, then $|f(z)| \leq |z|$ for all $z$ in the disk. Furthermore, $|f'(0)| \leq 1$, and if $|f(z)| = |z|$ for some $z$ in the disk or $|f'(0)| = 1$, then $f(z) = \alpha z$ for some $|\alpha| = 1$.
If $f$ is a holomorphic function on the unit disk such that $f(0) = 0$ and $|f(z)| < 1$ for all $z$ in the unit disk, then either $|f(z)| \leq |z|$ for all $z$ in the unit disk, or $|f(z)| = |z|$ for some $z$ in the unit disk.
If $d \cdot a \leq k \leq d \cdot b$, then there exists a point $c$ on the line segment from $a$ to $b$ such that $d \cdot c = k$ and $d \cdot z \leq k$ for all $z$ on the line segment from $a$ to $c$, and $k \leq d \cdot z$ for all $z$ on the line segment from $c$ to $b$.
If $f$ is holomorphic on a convex open set $S$ and continuous on $S$, then the integral of $f$ around any triangle in $S$ is zero.
If $f$ is holomorphic on the interior of a convex open set $S$ and continuous on $S$, then the integral of $f$ around any triangle in $S$ is zero.
If $f$ is a continuous function on a convex open set $S$ and $f$ is holomorphic on the two open sets $S_1$ and $S_2$ where $S = S_1 \cup S_2$, then the contour integral of $f$ along the boundary of $S$ is zero.
Suppose $S$ is a convex open set in $\mathbb{R}^n$, $a, b, c \in S$, $d \in \mathbb{R}^n$ is a nonzero vector, and $f$ is a continuous function on $S$ that is holomorphic on the two sets $\{z \in S : d \cdot z < k\}$ and $\{z \in S : k < d \cdot z\}$ for some $k \in \mathbb{R}$. Then the contour integral of $f$ along t...
If $f$ is holomorphic on the upper and lower half-planes and continuous on the real line, then $f$ is holomorphic on the whole plane.
If $f$ is holomorphic on the upper half-plane and continuous on the real line, then the Schwarz reflection of $f$ is holomorphic on the whole plane.
If $f$ is a holomorphic function on a ball of radius $r$ centered at $0$, and $f(0) = 0$, and the derivative of $f$ is bounded by $2$ times the derivative of $f$ at $0$, then the image of the ball of radius $r$ under $f$ contains a ball of radius $(3 - 2 \sqrt{2}) r \|f'(0)\|$.
If $f$ is holomorphic on a ball $B(a,r)$ and the norm of the derivative of $f$ is bounded by $2$ times the norm of the derivative of $f$ at $a$, then the image of $B(a,r)$ under $f$ contains a ball of radius $(3 - 2 \sqrt{2}) r \|f'(a)\|$.
If $f$ is holomorphic on the unit ball around $a$ and $f'(a) = 1$, then there exists a ball $B$ of radius at least $1/12$ such that $f(B)$ contains the unit ball around $f(a)$.
If $f$ is holomorphic on a ball of radius $r$ centered at $a$, then $f$ maps a ball of radius $r' \leq r \frac{\|f'(a)\|}{12}$ centered at $a$ onto a ball of radius $r'$.
Suppose $f$ is holomorphic on a set $S$ and $a \in S$. If $t$ is a positive real number such that $t \leq \text{dist}(a, z)$ for all $z \in \partial S$, then there exists a ball $B$ of radius $r \leq t \cdot \|f'(a)\| / 12$ such that $B \subseteq f(S)$.
A set $S$ is connected if and only if it is not the union of two nonempty open sets that are disjoint.
For any property $P$ of sets, the statement $\exists S. P(-S)$ is equivalent to the statement $\exists S. P(S)$.
A set $S$ is connected if and only if every clopen subset of $S$ is either empty or $S$ itself.
If $T$ is a connected subset of $S$ and $x, y \in T$, then $x$ and $y$ are in the same connected component of $S$.
If $x$ and $y$ are in the same connected component of $S$, then $x$ and $y$ are in $S$.
The connected component of a point $x$ in a set $S$ containing $x$ is $x$ itself.
The connected component of $x$ in $S$ is $x$ if and only if $x \<in> S$.
If $x$ and $y$ are in the same connected component of $S$, then $y$ and $x$ are in the same connected component of $S$.
If $x$ and $y$ are in the same connected component of $S$, and $y$ and $z$ are in the same connected component of $S$, then $x$ and $z$ are in the same connected component of $S$.
If $x$ and $y$ are in the same connected component of $S$, and $S$ is a subset of $T$, then $x$ and $y$ are in the same connected component of $T$.
The connected component of $x$ in $S$ is the union of all connected subsets of $S$ that contain $x$.
The connected component of a point $x$ in a topological space $S$ is connected.
A set $S$ is connected if and only if for every $x \in S$, the connected component of $x$ in $S$ is $S$ itself.
The connected component of a point $x$ in a topological space $S$ is a subset of $S$.
If $S$ is a connected set and $x \in S$, then the connected component of $x$ in $S$ is $S$ itself.
A topological space is connected if and only if any two points in the space are in the same connected component.
If $T$ is a connected subset of $S$ containing $x$, then $T$ is contained in the connected component of $S$ containing $x$.
If $S \subseteq T$, then the connected component of $x$ in $S$ is a subset of the connected component of $x$ in $T$.
The connected component of $x$ in $S$ is empty if and only if $x$ is not in $S$.
The connected component of the empty set is the empty set.
If $y$ is in the connected component of $x$, then the connected component of $y$ is the same as the connected component of $x$.
If $S$ is a closed set, then the connected component of $S$ containing $x$ is closed.
The connected component of $a$ is disjoint from the connected component of $b$ if and only if $a$ is not in the connected component of $b$.
The connected components of a set $S$ are pairwise disjoint if and only if they are distinct.
The intersection of two connected components is nonempty if and only if they are the same connected component.
The connected component of $x$ in $S$ is the same as the connected component of $y$ in $S$.
The connected component of $x$ is equal to the connected component of $y$ if and only if $x$ and $y$ are either both not in $S$ or both in $S$ and $x$ and $y$ are in the same connected component of $S$.
A set $S$ is connected if and only if all of its connected components are equal.
The connected component of a point $x$ in a set $S$ is the same as the connected component of $x$ in the connected component of $x$ in $S$.
If $x$ is in a connected set $c$ and $c$ is contained in $S$, and if $c$ is the only connected set containing $x$ that is contained in $S$, then $c$ is the connected component of $x$ in $S$.
If $T$ is a connected subset of a topological space $S$, and $x$ and $y$ are points of $T$ that belong to the same connected component of $S$, then $x$ and $y$ belong to the same connected component of $T$.
The union of all connected components of a set $S$ is $S$.
The complement of a connected component of a set $S$ is the union of all the other connected components of $S$.
If $T$ is a subset of $U$ and $a$ is a point in $U$, then the connected component of $a$ in $T$ is the same as the connected component of $a$ in $U$.
A set $S$ is a component of $U$ if and only if there exists $x \in U$ such that $S$ is the connected component of $U$ containing $x$.
If $x \in U$, then the connected component of $x$ in $U$ is a component of $U$.
If $S$ is a component of $U$, then there exists $x \in U$ such that $S$ is the connected component of $U$ containing $x$.
The union of the components of a topological space is the space itself.
The components of a topological space are pairwise disjoint.
The components of a topological space are nonempty.
If $c$ is a component of $s$, then $c$ is a subset of $s$.
If $c$ is a component of $s$, then $c$ is connected.
A set $c$ is a component of $s$ if and only if $c$ is nonempty, $c$ is a subset of $s$, $c$ is connected, and $c$ is maximal with respect to these properties.
If $t$ is a connected subset of $s$ and $c_1$ and $c_2$ are components of $s$ that intersect $t$, then $c_1 = c_2$.
If $s$ is a closed set and $c$ is a component of $s$, then $c$ is closed.
If $c$ and $c'$ are components of $s$, then $c$ and $c'$ are disjoint if and only if $c \neq c'$.
Two components of a topological space are equal if and only if they intersect.
The components of a set are empty if and only if the set is empty.
The components of the empty set are the empty set.
A set is connected if and only if it has exactly one connected component.
A set $s$ is connected if and only if the set of components of $s$ is equal to $\{s\}$.
A set $s$ has exactly one component if and only if $s$ is connected and nonempty.
A set is connected if and only if it has no proper connected subsets.
A set is connected if and only if all of its components are equal.
A set $s$ is a component of itself if and only if it is connected and nonempty.
If $c$ is a component of $s$, $t$ is a connected subset of $s$, and $t$ intersects $c$, then $t$ is contained in $c$.
If $t$ is a connected subset of $s$ and $s$ is nonempty, then there exists a component of $s$ that contains $t$.
If $s$ is a component of $u$ and $s \subseteq t \subseteq u$, then $s$ is a component of $t$.
If $c$ is a component of $s$, then $s - c$ is the union of all components of $s$ other than $c$.
If $s$ is connected and $t$ is contained in the closure of $s$, then $t$ is connected.
The connected component of a point $x$ in a set $s$ is closed in the subspace topology on $s$.
If $C$ is a component of a set $s$, then $C$ is closed in the topology of $s$.
If $f$ is a continuous function on a connected set $S$, and the level set $\{x \in S : f(x) = a\}$ is open in $S$, then either $f$ is never equal to $a$ on $S$, or $f$ is always equal to $a$ on $S$.
If $f$ is a continuous function on a connected set $S$, and the level set $\{x \in S \mid f(x) = a\}$ is open in $S$, then either $f$ is identically equal to $a$ on $S$, or $f$ is never equal to $a$ on $S$.
If $f$ is a continuous function on a connected set $S$ and $f$ has an open level set, then $f$ is constant.
If two topological spaces are homeomorphic, then they are connected if and only if they are connected.
If $f$ is a continuous function from a topological space $S$ to a connected topological space $T$, and if the preimage of every point in $T$ is connected, then $S$ is connected.
If $f$ is a continuous function from a set $S$ to a set $T$, and $f$ maps open sets in $S$ to open sets in $T$, and the preimage of any point in $T$ is connected, then the preimage of any connected set in $T$ is connected.
If $f$ is a continuous function from a set $S$ to a set $T$, and if $f$ maps closed sets in $S$ to closed sets in $T$, and if the preimage of every point in $T$ is connected, then the preimage of every connected set in $T$ is connected.
If $S$ and $U$ are connected sets with $S \subseteq U$, and $T$ is a clopen subset of $U - S$, then $S \cup T$ is connected.
If $S$ and $U$ are connected sets with $S \subseteq U$, and $C$ is a component of $U - S$, then $U - C$ is connected.
If $f$ is a continuous function from a connected set $S$ to a set $t$ such that for every $y \in t$, the connected component of $y$ in $t$ is just $\{y\}$, then $f$ is constant.
If every continuous function from a set $S$ to a normed vector space $V$ with finite range is constant, then $S$ is connected.
If $f$ and $g$ are differentiable at $z$, $f(z) = g(z) = 0$, $g'(z) \neq 0$, and $\frac{f'(z)}{g'(z)} = c$, then $\lim_{w \to z} \frac{f(w)}{g(w)} = c$.
If a function is not integrable, then its integral is zero.
If $f$ has a contour integral $i$ along $g$, then the contour integral of $f$ along $g$ is $i$.
If $f$ has a contour integral along a path $p$ and $f$ is contour integrable along a path $\gamma$, and the contour integral of $f$ along $p$ is equal to the contour integral of $f$ along $\gamma$, then $f$ has a contour integral along $\gamma$.
If $f$ is integrable on a contour $i$, then $f$ has a contour integral on $i$ equal to the value of the contour integral of $f$ on $i$.
If a function $f$ has a contour integral $i$ along a contour $g$, and $f$ has a contour integral $j$ along the same contour $g$, then $i = j$.
If a function $f$ has a contour integral along a contour $g$, then $f$ is contour integrable along $g$.
The integral of $f(g(x)) \cdot g'(x)$ over $[a,b]$ is the same as the integral of $f(g(x)) \cdot g'(x)$ over $[a,b]$ with respect to the localized vector derivative.
The integral of $f(g(x)) \cdot g'(x)$ is the same as the integral of $f(g(x)) \cdot g'(x)$ over the interval $[a,b]$.
The contour integral of $f$ along a curve $g$ is equal to the integral of $f \circ g$ times the derivative of $g$.
A function $f$ is contour-integrable on a path $g$ if and only if the function $t \mapsto f(g(t)) \cdot g'(t)$ is integrable on the interval $[0,1]$.
If $f$ has a contour integral along a path $g$, then $f$ has a contour integral along the reverse path $g^{-1}$ with the opposite sign.