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Let $q:X\to X/\mathcal{R}$ send each point to its equivalence class. The quotient topology isIt is the finest topology on the quotient set making the canonical surjection $q$ continuous. The preimage test, rather than a choice of representatives, defines its open sets.

The scalar-valued multivariable mean value theorem states that, if $f$ is differentiable on an open set containing the closed line segment from $x$ to $y$, then for $x\ne y$ there exists $s\in (0,1)$ such thatThis follows by applying the one-dimensional mean value theorem to $t\mapsto f(x+t(y-x))$ on $[0,1]$. The segment condition is necessary for this formulation on a possibly nonconvex domain.

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

A basis of a topology $\mathcal{T}$ is a collection $\mathcal{B}\subseteq \mathcal{T}$ such that every open set is a union of members of $\mathcal{B}$. Equivalently, for each open $U$ and every $x\in U$, there exists $B\in \mathcal{B}$ with $x\in B\subseteq U$. The empty union represents the empty set. In particular, the basis covers $X$, and any intersection of two basis elements is itself a union of basis elements.

Yes: at least one factor topology must be indiscrete. Suppose instead that both have a nonempty proper open set, say $\varnothing \ne U⊊X$ and $\varnothing \ne V⊊Y$. Thenis open in the product topology. Both coordinate projections of $W$ are the full factors, because $U,V$ are nonempty. If $W$ were a rectangle $A\times B$, this would force $A=X$, $B=Y$, hence $W=X\times Y$. But choosing $x\notin U$ and $y\notin V$ gives $(x,y)\notin W$, a contradiction. Thus both factors cannot have nontrivial open sets. This is rectangularity of every product-open set.

Let $q:X\to X/\mathcal{R}$ send each point to its equivalence class. The quotient topology isIt is the finest topology on the quotient set making the canonical surjection $q$ continuous. The preimage test, rather than a choice of representatives, defines its open sets.

The scalar-valued multivariable mean value theorem states that, if $f$ is differentiable on an open set containing the closed line segment from $x$ to $y$, then for $x\ne y$ there exists $s\in (0,1)$ such thatThis follows by applying the one-dimensional mean value theorem to $t\mapsto f(x+t(y-x))$ on $[0,1]$. The segment condition is necessary for this formulation on a possibly nonconvex domain.

The map $G$ is $C^1$ on the open ball $B$. Its Jacobian matrix isAt the origin $G(0)=0$ and $\boxed{\det DG(0)=1/3\ne 0}$. The inverse function theorem therefore supplies the required open sets $V_1\subset B$ and $V_2\subset {\mathbb{R}}^3$, both containing zero, with $G:V_1\to V_2$ a diffeomorphism. The explicit inverse formula on its image,also checks the local conclusion directly.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open image, with continuously differentiable inverse. Equivalently one may specify a one-to-one continuously differentiable map whose Jacobian determinant is everywhere nonzero; the inverse function theorem gives the corresponding smooth inverse on its image.

False. Let $X$ be an indiscrete two-point space and $Y$ a one-point space. The constant map is continuous and onto; $X$ is compact because it is finite and connected because its only nonempty open set is the whole space. It is not Hausdorff, since its two points have no disjoint nonempty neighbourhoods. The one-point codomain is Hausdorff.

The function is continuous if it is continuous at every point of $X$. Equivalently, the preimage of each open set in $Y$ is open in $X$.

Every point of $A$ differs from $a$, so its singleton is an open ball by part (ii). The union of these singleton open sets is $A$. Thus $A$ is open whenever $a\notin A$.

A separation of $f(X)$ would pull back under the continuous map to two disjoint nonempty open sets covering $X$: nonemptiness follows because their points belong to the image. This contradicts connectedness of $X$. Hence a continuous image of a connected space is connected.

A subset is connected when its subspace topology admits no separation into two disjoint nonempty open sets. Equivalently, its only subsets that are both open and closed in that subspace are the empty set and the entire subset.

The difference map $(z,w)\mapsto z-w$ is continuous, and $X$ is its preimage of the open set $\mathbb{C}\setminus \{0\}$. Thus $X$ is open in the product.

The class of $(z,w)$ is $C_z=\{z\}\times (\mathbb{C}\setminus \{z\})$. A punctured plane is path-connected: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is connected.

The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is continuous as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-topology basis gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq \pi (U)$. Hence $\pi (U)$ is open.

The map is constant exactly on the equivalence classes, so it induces a bijection $\overline{\pi }:X/R\to \mathbb{C}$. It is continuous by the defining property of the quotient topology. It is also open: for quotient-open $V$, $\overline{\pi }(V)=\pi (q^{-1}(V))$ is open. ThusThe plane is connected, and the classes are connected, so part (b) now yields $X$ is connected.

Let $C$ be a component of the open set $X\subseteq \mathbb{R}$. Being a connected subset of the real line, $C$ is an interval. For any $x\in C$, openness of $X$ gives $ϵ>0$ with $(x-ϵ,x+ϵ)\subseteq X$. This interval is connected and meets $C$, so their union is connected. Maximality of $C$ makes the entire interval lie in $C$. Thus every point of $C$ is interior, so each component is an open interval, allowing unbounded endpoints.

Every point belongs to its component, so the components cover $X$; by part (ii) they are pairwise disjoint, and by part (iii) they are nonempty open intervals. Each contains a rational number. Fix an enumeration of $\mathbb{Q}$ and choose the first rational belonging to each component. Distinct components receive distinct rationals, giving an injection from the component family into a countable set. ThereforeThe empty open set is represented by the empty union.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

A connected space has no expression as the union of two disjoint nonempty open subsets. A continuous surjection to the two-point discrete space would pull back its singleton sets to such a separation. Conversely, a separation $X=U\bigsqcup V$ defines a continuous surjection by taking the values $0$ on $U$ and $1$ on $V$. Thus these two conditions are equivalent, including for the empty space.

If a subset $S\subseteq \mathbb{R}$ contains $a<c$ but omits some $b$ between them, its intersections with $(-\infty ,b)$ and $(b,\infty )$ separate it. Hence every nonempty connected subset is an interval. Conversely, let an interval $I$ have a separation and choose $a<b$ in its opposite sides. Restrict the associated two-valued continuous function to $[a,b]\subseteq I$ and regard it as a real-valued function. The intermediate value theorem forces the value $1/2$, a contradiction. Singletons and the empty set are connected directly.

For the union assertion, let $u:X\to \{0,1\}$ be continuous. Each restriction $u{|}_{A_i}$ is constant because $A_i$ is connected. Since every pair of the $A_i$ intersects, these constants agree, so $u$ is constant on $X$. There is no continuous surjection, and $X$ is connected. The empty indexing case gives $X=\varnothing$ and is harmless.

To apply this to the plane, for each $t\in \mathbb{R}$ take the union of the horizontal line at height $t$ with the vertical axis. Its two lines are connected and intersect; all such unions share the vertical axis and cover ${\mathbb{R}}^2$. Thus the plane is connected. For the punctured plane, take, for every $r>0$, the circle of radius $r$ together with the positive horizontal ray. The circle is the continuous image of $[0,2\pi ]$, and intersects that connected ray. These unions share the ray and cover ${\mathbb{R}}^2\setminus \{0\}$, so the punctured plane is connected too. Translation gives the same conclusion after removing any one point.

Finally, suppose $g:{\mathbb{R}}^2\to \mathbb{R}$ were a continuous injection. Its image $I$ is a connected interval with at least three points. Choose an interior point $c$ of $I$ and its unique preimage $p$. The restriction of $g$ to ${\mathbb{R}}^2\setminus \{p\}$ is a continuous surjection onto $I\setminus \{c\}$. Its domain is connected, whereas its image is separated by the two half-lines on either side of $c$. This contradicts preservation of connectedness under continuous images. Therefore no continuous injection of the plane into the line exists.

If $x\notin \overline{A}$, some closed superset $F$ of $A$ misses $x$. Its complement is an open set containing $x$ and disjoint from $A$. Conversely, if an open set $U$ contains $x$ and misses $A$, the closed set $X\setminus U$ contains $A$ and misses $x$, so $x\notin \overline{A}$.

Negating this equivalence gives the neighbourhood characterization