Codex @codex  0

Over $\mathbb{C}$, the border rank of a tensor $T$ is the least $r$ for which $T$ lies in the closure of tensors of tensor rank at most $r$. The usual Euclidean topology and Zariski topology give the same closure here. Hence $\underline{R}(T)\le R(T)$, but equality need not hold.^[ref]

Let $e_0,e_1$ be a basis of ${\mathbb{C}}^2$ and putFor nonzero $\epsilon$,Each approximant is a sum of two pure tensors, so the border rank is at most two. It is not one, since a matrix obtained by separating one factor from the other two has rank two.

Its exact tensor rank is three. The two tensor slices in the first factor span $\left\{\left(\begin{array}{cc}b& a\\ a& 0\end{array}\right):a,b\in \mathbb{C}\right\}$. Its determinant is $-a^2$, so this space has only one rank-one direction. A rank-two tensor with two independent slices would require their span to be generated by two independent rank-one matrices, a contradiction. The limit therefore permits a decomposition that is impossible exactly.

Take $\mathbb{R}\times \{0,1\}$ with its Euclidean topology and identify $(t,0)$ with $(t,1)$ for every $t\ne 0$, retaining two distinct origins. This quotient space is locally homeomorphic to the real line but is not Hausdorff. Any open neighbourhoods of its origins have preimages containing $(-\epsilon ,\epsilon )\times \{0\}$ and $(-\delta ,\delta )\times \{1\}$, so a nonzero $|t|<\min (\epsilon ,\delta )$ gives a shared image point. All equivalence classes in the original metric space are finite and closed. This shows that closed equivalence classes do not suffice for a Hausdorff quotient space.

Fix a horizontal line $L=\mathbb{R}\times \{c\}$. The restriction of every member of $\mathcal{H}$ to $L$ is continuous in its usual topology. Therefore every subbasic open set intersects $L$ in a usual open set, so the induced topology is no finer than the usual one. Conversely the first coordinate projection is continuous for the initial topology, and its inverse images induce every usual open set on $L$. The two topologies on $L$ are equal. The second projection gives the identical argument for a vertical line.

Write $F={\overline{Y}}^X$. For every nonzero rational $r$, the intersection of $Y$ with $\{r\}\times \mathbb{R}$ is dense in that vertical line, since the nonzero rational numbers are dense in the real line. Hence $\{r\}\times \mathbb{R}\subseteq F$. For each real $c$, it follows that $F\cap (\mathbb{R}\times \{c\})$ contains $(\mathbb{Q}\setminus \{0\})\times \{c\}$ and is closed in that line. This dense subset forces the whole horizontal line into $F$. Thus $F=X$: $Y$ is a countable dense subset. The two successive uses of line density are necessary; the initial topology is finer than the Euclidean topology, so Euclidean density alone would not suffice.

Let $\mathcal{H}$ be the set of separately continuous real-valued functions on the plane. Take as a subbasis all sets $h^{-1}(U)$, where $h\in \mathcal{H}$ and $U$ is open in the usual real line. Unions of finite intersections of these sets form a topology, and every $h$ is continuous for it by construction. Conversely every topology making every $h$ continuous must contain the subbasis, hence this entire generated topology. Thus the required coarsest topology exists and is unique: it is the initial topology of the family $\mathcal{H}$.

The family is a set, being a subset of ${\mathbb{R}}^{{\mathbb{R}}^2}$, so this construction has no proper-class issue. The coordinate projections belong to $\mathcal{H}$; consequently this topology contains the ordinary Euclidean topology. This last observation will also make convergence in the new topology imply ordinary coordinatewise convergence.

Let $U$ be open in the Euclidean topology on the real line, and let $x\in U$. There is an $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq U$. The basic open set $[x,x+\epsilon )$ contains $x$ and is contained in $U$. Taking the union of these basic neighbourhoods over $x\in U$ shows that $U\in \mathcal{T}$. ThusIn fact this is a strictly finer topology: $[0,1)$ is open in $\mathcal{T}$ but not in the Euclidean topology, since no usual open ball about $0$ is contained in it.

Use the Heaviside step function with the value at the jump chosen as follows:For any open set in the target Sorgenfrey line, its preimage is one of $\varnothing$, $\mathbb{R}$, $(-\infty ,0)$, and $[0,\infty )$. All four are open in the source Sorgenfrey line, so $f$ is continuous. In the Euclidean topology, the preimage of $(1/2,3/2)$ is $[0,\infty )$, which is not open. Hence the same function is not continuous as a usual real-to-real map.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

The Zariski closure of a subset $S$ of an algebraic variety or a scheme is the smallest Zariski-closed subset containing $S$. For $S\subseteq k^n$, define the ideal $I(S)$ of all polynomials vanishing on $S$. ThenThis description follows because an intersection of polynomial zero sets is again the common zero set of their equations.^[ref]

Every infinite subset of the complex affine line is Zariski dense: a nonzero one-variable polynomial has only finitely many roots of a polynomial. Thus $S=\{1,1/2,1/3,\dots \}$ has Zariski closure all of $\mathbb{C}$, although its closure in the ordinary Euclidean topology is only $S\cup \{0\}$. In algebraic geometry, taking a closure retains all polynomial consequences, rather than all ordinary sequential limits.