''simple arc, Jordan arc''
An intrinsic characterization is: A simple arc is a [[Line (curve)|line (curve)]] that has ramification index 1 at two poi
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along a non-closed simple arc $ \gamma = \{ {z ( t) } : {0 \leq t \leq 1 } \} $''
The removal of the points of the arc $ \gamma $
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...$ is non-trivial for the arc $L_1$ (Fig. a); this group is trivial for the arc $L_2$ (Fig. b), but $E^3\setminus L_2$ itself is not homeomorphic to the co
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...]] in which any two points can be joined by a continuous image of a simple arc; that is, a space $ X $
...ace is a Hausdorff space in which any two points can be joined by a simple arc, or (what amounts to the same thing) a Hausdorff space into which any mappi
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...Cantor curve have the same finite branch index, then the Cantor curve is a simple closed line. The universal Cantor curve (the Menger curve) can be construct
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...any loop in $V\setminus qM$ is homotopic to zero in $U\setminus qM$ (local simple connectedness). If $m=n-2$, then such a criterion holds for $n\neq4$, but i
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...the least upper bound of the lengths of the broken lines inscribed in this arc. Any continuous curve has a length, finite or infinite. If its length is fi
Then the length of an arc of the curve counted from the point corresponding to the parameter value $
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be the end point of the arc on the unit circle $ x ^ {2} + y ^ {2} = 1 $(
The arc from $ B $
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...2$ be a quadratic differential on $R$; let $C$ be the set of all zeros and simple poles of $Q(z)\,dz^2$ and let $H$ be the set of all poles of $Q(z)\,dz^2$ o
...C\cup H)$ there passes a trajectory of $Q(z)\,dz^2$ that is either an open arc or a Jordan curve on $R$.
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are contained in a simple arc lying in $ K $.
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...l curvature of the surface on the set of points of the curve. For a simple arc $ L $,
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...Jordan [[#References|[1]]]. Together with the similar assertion: A simple arc does not decompose the plane, this is the oldest theorem in set-theoretic t
...ounded component $A$ and every point $x_0\in\Gamma$, there exists a simple arc with ends $x_0$ and $x$ and all points of which, except $x_0$, are containe
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...nd have the property that every proper subcontinuum is an [[Arc (topology)|arc]]. See [[#References|[a7]]].
...sed curve is the only homogeneous bounded plane continuum that contains an arc" ''Canad. Math. J.'' , '''12''' (1960) pp. 209–230</TD></TR><TR><TD va
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...cible continuum. Every locally connected irreducible continuum is a simple arc, that is, is homeomorphic to an interval.
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...the same direction) and such edges (arcs) are then said to be multiple. An arc (or edge) can begin and end at the same vertex, in which case it is known a
...ne says that an edge $\{u,v\}$ connects two vertices $u$ and $v$, while an arc $(u,v)$ begins at the vertex $u$ and ends at the vertex $v$. Each graph can
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is defined as any simple Jordan arc $ c = \overline{ {ab }}\; $,
and such that the arc $ c $
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...the line" (N.N. Luzin). There are no Peano curves at which every point is simple or two-fold, but there is a Peano curve with as multiple points (only count
...)$, $z=t$, where the first two functions give a Peano curve. Although this arc is impermeable to rain, it is by no means a continuous surface.
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The following are some very simple solutions:
...nside $\gamma$ and $D$ is the domain bounded by a part of $\gamma$ and the arc $\Gamma$, with $0 \notin \overline { D }$, then for $z \in D$ and $f \in H
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...there is a disc $D$ embedded in $M$ such that $D\cap F=\partial D$ and the simple closed curve $\partial D$ does not bound a disc in $F$. Otherwise, such a s
...\cap\partial M=\beta$, and $\alpha$ does not cobound a disc in $F$ with an arc in $\partial F$. If a properly embedded surface $F$ in a three-dimensional
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plane, having as its only singularity a simple pole at the point $ s = 1 $
by a circular arc of radius $ R = 2 N + 1 $
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