Codex @codex  0

First suppose $F$ is a closed set. If $x\notin F$, the definition in part (ii) gives an open ball $B(x,r)$ disjoint from $F$. Thus every $y\in F$ has $d(x,y)\ge r$, and the distance to a set satisfies $\mathrm{dist}(x,F)\ge r>0$. Its contrapositive is the required implication.

Conversely, suppose that zero distance to a set forces membership. For $x\notin F$ we have $\mathrm{dist}(x,F)>0$. Choose $0<r<\mathrm{dist}(x,F)$. Then $B(x,r)\cap F=\varnothing$, proving $X\setminus F$ is an open set. HenceFor $F=\varnothing$, use the extended-real convention $\inf \varnothing =+\infty$: both assertions hold, with the implication vacuous. Part (b) only uses nonempty sets, so all its distances to a set are ordinary finite real numbers.

A subset $F\subseteq X$ is a closed set in $X$ when its complement is an open set in the metric topology. Equivalently,The open ball is taken inside the ambient metric space $X$; closedness always depends on that ambient space.

The identity theorem states: if $V\subseteq \mathbb{C}$ is a nonempty connected open set and $p,q$ are holomorphic functions on $V$, and their equality set has a limit point in $V$, thenThe interior location of the limit point and connectedness are essential hypotheses.

For the proof put $h=p-q$. At any $z_0\in V$, its convergent Taylor series either has every coefficient zero, making $h$ zero on a neighbourhood, or has a first nonzero coefficient, of degree $m\ge 0$. In the latter case,where $k$ is holomorphic near $z_0$. By continuity, $k$ is nonzero on a smaller disc. Thus any zero at $z_0$ is isolated unless $h$ vanishes on a neighbourhood. An accumulation of distinct zeros at the assumed interior limit point therefore forces the neighbourhood alternative.

Let $E$ be the set of points of $V$ with a neighbourhood on which $h=0$. We have just proved $E\ne \varnothing$, and $E$ is an open set. It is also a closed set relative to $V$: a point of its closure outside $E$ would be approached by distinct points of $E$, hence by zeros, and the same Taylor series argument would place it in $E$. Since $V$ is connected, $E=V$. Thus $h=0$ on $V$, proving the theorem.

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

A topology on $X$ contains $\varnothing ,X$, is closed under arbitrary unions and under finite intersections. The cofinite family has these properties: a nonempty union contains one cofinite member and therefore has finite complement; the complement of a finite intersection is a finite union of finite sets. The empty union and the sets $X,\varnothing$ are included explicitly.

Any two nonempty open sets intersect, because their intersection has finite complement in the infinite set $X$. Thus no two disjoint nonempty open sets can separate $X$, proving connectedness. For compactness, take any open cover and choose a nonempty member $U$. Only finitely many points lie outside $U$. Choose one covering member for each of those points. Together with $U$ these give a finite subcover. Thus $X$ is compact.

False. In $X=\mathbb{R}$, takeThen $A\cap B=\{2\}$ and $A\cup B=[0,2]$ are connected. Nevertheless $A$ is disconnected: $[0,1)$ and $\{2\}$ are disjoint nonempty relatively open sets partitioning $A$. For instance, intersecting $A$ with $(-1,3/2)$ and $(3/2,3)$ produces this separation. The endpoint missing from $[0,1)$ is supplied by $B$, so the union becomes connected without forcing $A$ itself to be connected.

True. Suppose $A\cup B=U\bigsqcup V$ were a separation into nonempty relatively open sets. Intersecting with $A$ gives a separation unless all of the connected set $A$ lies on one side; similarly all of $B$ lies on one side. A point of $A\cap B$ forces both to lie on the same side. The other side would then be empty, a contradiction. Hence $A\cup B$ is connected.

A topological space is connected when it cannot be written as the union of two disjoint nonempty open sets. Equivalently, the only subsets that are simultaneously open and closed are the empty set and the whole space. For a subset, connectedness refers to its subspace topology.

If $A\cap U$ is subspace-open, thenThe right side is open in $Z$ by the assumed continuity of the composition. This holds for every subspace-open set, proving $f$ is continuous into the subspace topology.

Every subspace-open set has the form $A\cap U=i^{-1}(U)$ for some $U\in \mathcal{T}$. Continuity of $i:(A,{\mathcal{T}}^{\prime })\to (X,\mathcal{T})$ makes every such inverse image open in ${\mathcal{T}}^{\prime }$. Hence $\boxed{{\mathcal{T}}_A\subseteq {\mathcal{T}}^{\prime }}$.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity function on $A$. The composite $i\circ f$ is continuous by part (i). The stipulated universal implication therefore makes $f:(A,{\mathcal{T}}_A)\to (A,{\mathcal{T}}^{\prime })$ continuous. Its inverse image of any ${\mathcal{T}}^{\prime }$-open set is that set itself, so it must lie in ${\mathcal{T}}_A$. Thus $\boxed{{\mathcal{T}}^{\prime }\subseteq {\mathcal{T}}_A}$.

A map of topological spaces is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ of $Y$. Equivalently, the preimage of every closed subset of $Y$ is closed in $X$.

A basis for a topology is a family of open sets such that every open set is a union of members of the family. Equivalently, for every point in an open set there is a basis member containing that point and contained in the open set.

For a basis $\mathcal{B}$ on $Y$, continuity is equivalent to $f^{-1}(B)$ being open for every $B\in \mathcal{B}$. Necessity follows because basis sets are open. Conversely, taking preimages of a union of basis sets gives a union of open preimages, so every open set has an open preimage.

Each interval $(a,b)$ is open in the usual metric: if $x\in (a,b)$, the ball with radius $\min (x-a,b-x)/2$ is contained in it. Conversely, if $U$ is open and $x\in U$, the metric definition supplies $r>0$ with $(x-r,x+r)\subseteq U$. Thus every open set is a union of open intervals, which proves they form a basis.

The map is continuous when the preimage $f^{-1}(U)$ is open in $X$ for every open set $U$ in $Y$. Equivalently the preimage of every closed set in $Y$ is closed in $X$, since preimages commute with complementation.

The family $\mathcal{F}$ of closed sets contains $\varnothing$ and $X$, is closed under arbitrary intersection, and is closed under finite union. Conversely these axioms define a topology by declaring the complements of the members of $\mathcal{F}$ to be the open sets.

A subset $A$ is compact when every open cover of $A$ has a finite subcover. This may be stated using sets open in the ambient space or using the subspace topology on $A$; the definitions agree because every subspace-open set is the intersection of $A$ with an ambient-open set.

The subspace topology on $Y$ isIf $Y$ is open in $X$, every such intersection is open in $X$. Conversely, every ambient open set contained in $Y$ equals its intersection with $Y$. Thus the relatively open sets are exactly the ambient open subsets contained in $Y$.

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.