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A metric space in which every two points are joined by a metric geodesic segment. Equivalently, they admit a continuous path whose metric path length equals their distance. In this general setting no Riemannian metric or covariant derivative is required.

A continuous path with finite arc length on every compact parameter interval. Families of such paths are used in the measurable-density definition of extremal length.

For a continuous path $\gamma :[a,b]\to X$ in a metric space, take the supremum over finite partitions $a=t_0<\cdots <t_n=b$ of ${\sum }_{i=1}^{n}d(\gamma (t_i),\gamma (t_{i-1}))$. This definition allows $+\infty$ and needs no derivative. A metric geodesic parametrized on $[a,b]$ has length $b-a$. In a Riemannian manifold, it agrees with the usual arc length integral for a smooth curve.

Let $\mathcal{A}$ be the closure of the polynomials in $C(K)$ for the uniform norm. It is a closed algebra, since products of uniformly convergent sequences converge uniformly on the compact set $K$.

Choose a point $W$ with $|W|>{\sup }_{z\in K}|z|$. The uniformly convergent geometric seriesshows that this reciprocal belongs to $\mathcal{A}$. Join $w$ to $W$ by a continuous path in $\mathbb{C}\setminus K$. Its image is compact and disjoint from $K$, so their distance $d$ is positive. Partition the path into finitely many steps shorter than $d/2$.

Suppose $1/(b-z)\in \mathcal{A}$ and $|a-b|<d/2$, with $b$ on the path. Thenconverges uniformly on $K$, since $|b-z|\ge d$. Every partial sum belongs to $\mathcal{A}$, by its algebra property, and so does the limit. Working backwards along the partition transfers the pole from $W$ to $w$. Thus $1/(w-z)\in \mathcal{A}$: a polynomial $P$ exists withThis pole-moving polynomial approximation is the elementary mechanism behind the polynomial Runge theorem.

A standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ satisfying $B_0=0$ almost surely, having continuous paths almost surely, and having independent increments withHere $N(0,t-s)$ is the normal distribution with zero expected value and variance $t-s$. Thus the increments are also stationary increments.

If Brownian motion is defined relative to a specified filtration $({\mathcal{F}}_t)$, additionally require that it is an adapted process and that $B_t-B_s$ is independent of ${\mathcal{F}}_s$. For its natural filtration, this follows from independent increments. These are the defining properties of standard Brownian motion.

Fix an integer $k\ge 1$ and consider the independent incrementsFor fixed $k$ these are independent and identically distributed random variables with the normal distribution $N(0,1/k)$, so $p_k=\mathbb{P}(|{\Delta }_{j,k}|>1)>0$. The second Borel-Cantelli lemma gives $|{\Delta }_{j,k}|>1$ for infinitely many $j$, almost surely. Intersecting these probability-one events over the countably many $k$ preserves probability one.

On any path in this intersection, uniform continuity on $[0,\infty )$ would give some $\delta >0$ such that $|B_t-B_s|<1$ whenever $|t-s|<\delta$. Choosing $k$ with $1/k<\delta$ contradicts the existence of an increment $|{\Delta }_{j,k}|>1$. Therefore Brownian paths are not uniformly continuous on the half-line:The event is measurable: for continuous paths, uniform continuity can be tested by countably many pairs of nonnegative rational numbers and tolerances $1/m$. Brownian motion has uniform continuity on every compact time interval, but almost surely fails it on the whole half-line.

For each $k$, the coordinate projectionis continuous when $\mathbb{Z}$ has the discrete topology: if $d(x,y)<2^{-k}$, then $x_k=y_k$.

If $\gamma :[0,1]\to X$ were a continuous path, then ${\pi }_k\circ \gamma$ would be a continuous map from the connected interval $[0,1]$ to a discrete space, so it would be constant. This holds for every $k$, and therefore $\gamma$ itself must be constant. Since $X$ contains distinct points, no path joins every pair. Henceindeed, every path component is a singleton.

The implicit function theorem states that if $F(x_0,y_0)=0$ and the partial derivative $D_{y}F(x_0,y_0)$ is an invertible linear map, then near $(x_0,y_0)$ the zero set of $F$ is uniquely the graph $y=g(x)$ of a continuously differentiable function.

If $f$ is a differentiable bijection with differentiable inverse $g$, the chain rule applied to $g\circ f$ and $f\circ g$ givesThus $D{f}_x$ is an isomorphism with inverse $D{g}_{f(x)}$.

If a continuously differentiable map $F:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has invertible derivative everywhere, the inverse function theorem makes it a local diffeomorphism. In particular it is an open map, so its image is open. Its image need not be closed: $x\mapsto \arctan x$ has nonzero derivative everywhere and image $(-\pi /2,\pi /2)$.

For the given map of elementary symmetric polynomials,and direct evaluation of the determinant givesHence the critical set isIts complement is the Zariski-open set on which the three coordinates are pairwise distinct. Each point has one of the six possible strict coordinate orderings, and each ordering defines a nonempty convex open region. A continuous path cannot change an ordering without crossing $C$. Therefore ${\mathbb{R}}^3\setminus C$ has exactlyconnected components.

For a countable set $T\subset {\mathbb{R}}^2$, join two points outside $T$ through an auxiliary point avoiding the countably many lines through either endpoint and a point of $T$. Such an auxiliary point exists: choose a line with direction outside the countable list of forbidden directions, then avoid its countably many intersections with those lines. The two segments avoid $T$, giving a continuous path.