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An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

A Hausdorff space separates any two distinct points by disjoint open neighborhoods. For a subspace $A\subseteq X$, intersect such neighborhoods in $X$ with $A$ to obtain disjoint neighborhoods in the subspace topology. Thus subspaces inherit the Hausdorff property.

If $(z_1,z_2)$ lies off the diagonal of a Hausdorff space's square, choose disjoint open neighborhoods $U$ of $z_1$ and $V$ of $z_2$. The open rectangle $U\times V$ contains the pair and avoids the diagonal. Therefore its complement is open and the diagonal is closed.

The map $F:X\times Y\to Z\times Z$ given by $F(x,y)=(f(x),g(y))$ is continuous by the coordinate criterion from E3. The equality locus is $F^{-1}({\Delta }_Z)$ and is closed as an inverse image of a closed set.

Let $q:X\to X/\sim$ be the quotient map. If the quotient is Hausdorff, apply the same argument to $q$ in both coordinates:is closed in $X\times X$.

Conversely suppose $R$ is closed and $q$ is open. Take two distinct quotient points, with representatives $x,y$. Since $(x,y)\notin R$, an open rectangle $U\times V$ around it avoids $R$. The images $q(U)$ and $q(V)$ are open quotient neighborhoods. They are disjoint, because a quotient point in their intersection would have equivalent representatives $u\in U$, $v\in V$, giving $(u,v)\in R$. Hence the quotient is Hausdorff. Thus closedness of the relation and openness of the quotient map suffice for a Hausdorff quotient.

A subset is compact if every cover of it by open sets of the ambient space has a finite subcover; equivalently use open sets of its subspace topology. A cover of $A\cup B$ separately covers $A$ and $B$. Compactness of each supplies two finite collections, whose union is finite and covers $A\cup B$. If $A$ is closed in a compact space, add the open complement of $A$ to any open cover of $A$. Compactness supplies a finite subcover of the whole space, and discarding the complement leaves a finite cover of $A$.

For the enlarged space, every old open set is open in $C$ and has the same trace on $X$. The trace on $X$ of a new open set $V\cup \{\infty \}$ is $V$, which is open in $X$ because its complement is closed. Thus the induced topology on $X$ is exactly the original one.

Given an open cover of $C$, choose a member containing infinity. It has the form $V\cup \{\infty \}$ with $K=X\setminus V$ compact. The remaining cover restricted to $K$ has a finite subcover. Together with the chosen infinity-neighborhood, these finitely many sets cover $C$. Hence $C$ is compact, with no Hausdorff assumption needed.

Finally put $a=f(\infty )$. Continuity makes $U=f^{-1}((a-ϵ,a+ϵ))$ open and containing infinity. Therefore $U=V\cup \{\infty \}$ with $W=X\setminus V$ compact and closed in $X$. For every $x\in X\setminus W=V$ we haveThis expresses convergence to the value at infinity outside compact closed subsets.

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.

Let $C$ be any equivalence class. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the subspace topology on $C$. Its connectedness makes $C\cap A$ either empty or all of $C$. Thus an equivalence class meeting $A$ lies wholly in $A$. Hence $A$ is saturated, a union of whole equivalence classes.

A subset $A$ is connected if its subspace topology admits no separation into two disjoint nonempty relatively open subsets. Equivalently, every continuous map $A\to \{0,1\}$, with the target discrete, is constant: a nonconstant map gives the separation by its two fibers, and a separation defines such a map.

A subset $K$ of a topological space is compact if every cover of $K$ by open subsets of the ambient space has a finite subcover. This is equivalent to compactness of $K$ with its subspace topology.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

If the inclusion is continuous from $(A,{\mathcal{T}}^{\prime })$, then $i^{-1}(U)=A\cap U$ belongs to ${\mathcal{T}}^{\prime }$ for every $U\in \mathcal{T}$. These intersections are precisely all members of ${\mathcal{T}}_A$, soThus the subspace topology is the coarsest topology on $A$ making the inclusion continuous.

The subspace topology is ${\mathcal{T}}_A=\{A\cap U:U\in \mathcal{T}\}$. For every open $U\subseteq X$,The inverse-image definition of continuity therefore gives the inclusion is continuous from $(A,{\mathcal{T}}_A)$ to $(X,\mathcal{T})$.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity set map $A\to A$. The composition $i\circ f$ is the inclusion and is continuous by part (i). The hypothesized universal implication therefore says that the identity mapis continuous. Its preimage of a set $V\in {\mathcal{T}}^{\prime }$ is $V$ itself, so every such $V$ lies in ${\mathcal{T}}_A$. ThereforeCombining this with part (ii), when both properties hold, characterizes the subspace topology uniquely.

The subspace topology on $Y$ isIf $Y$ is open in $X$, each such intersection is open in $X$. Conversely, an open subset of $X$ contained in $Y$ is its own intersection with $Y$. Thus the open sets of $Y$ are exactly the open subsets of $X$ contained in $Y$ in this case.

We prove (i)$\Rightarrow$(ii). If $B=U\cap F$ with $U$ open and $F$ closed, then $\overline{B}\subseteq F$. Henceso $B=U\cap \overline{B}$. By the subspace topology definition this makes $B$ open in its closure.

An open cover of a topological space is a collection of open subsets whose union is the whole space. A compact space is one for which every open cover admits a finite subcover. The subspace topology on $A\subseteq Y$ consists of sets $A\cap O$ with $O$ open in $Y$.

Let $\{U_j\}$ cover $f(X)$ by relatively open sets, and write $U_j=f(X)\cap O_j$ with $O_j$ open in $Y$. By continuity, $\{f^{-1}(O_j)\}$ is an open cover of $X$. A finite subcollection covers $X$ by compactness, and its corresponding $U_j$ cover $f(X)$. Thus a continuous image of a compact space is compact.

For an image that is not closed, take $X$ to be a singleton and $Y=\{0,1\}$ with the indiscrete topology. Send the sole point to $0$. This map is continuous and $X$ is compact, while $f(X)=\{0\}$ is not closed in $Y$, whose only closed sets are $\varnothing$ and $Y$. A Hausdorff hypothesis on the target would exclude this example.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

A topological space is compact if every open cover has a finite subcover. It is connected if it cannot be written as the union of two disjoint nonempty open subsets. It is Hausdorff if every two distinct points have disjoint open neighborhoods. These definitions apply to the space itself and, with the subspace topology, to its subsets.

A subset $A$ is connected if, with its subspace topology, it cannot be written as $A=U\cup V$ with $U,V$ disjoint nonempty relatively open sets. Equivalently its only subsets both open and closed relative to $A$ are $\varnothing ,A$. This definition includes the empty set and singletons as connected sets.

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

Let $n=(0,0,1)$ and use the open coverThe cap $V$ is open in $Y$: it equals $Y\cap \{z>1/2\}$ because the attached chord has $z=0$. The cap is homeomorphic to an open disk, and $U\cap V=V\setminus \{n\}$ is a punctured open disk, hence homotopy equivalent to a circle. Both sets and their intersection are path-connected. Choose the base point $b$ in that intersection.

We first calculate ${\pi }_1(U,b)$. A stereographic projection identifies $S^2\setminus \{n\}$ with ${\mathbb{R}}^2$ and sends the two chord endpoints to $(-1,0),(1,0)$. Thus $U$ becomes the Euclidean plane with an extra closed interval attached at those two points. This description is a homeomorphism: gluing the two closed pieces along their endpoints gives exactly the original subspace topology.

The Euclidean plane has a deformation retraction onto $L=[-1,1]\times \{0\}$ while fixing its endpoints. An explicit homotopy isExtend it by the identity map on the extra chord. The pasting lemma for closed subspaces gives a deformation retraction of $U$ onto two distinct closed intervals sharing just their endpoints, a circle. Consequently ${\pi }_1(U,b)\cong \mathbb{Z}$, with the routine base point transport if the retraction moves $b$.

The inclusion-induced group homomorphism ${\pi }_1(U\cap V,b)\to {\pi }_1(U,b)$ is trivial: it factors through $S^2\setminus \{n\}\cong {\mathbb{R}}^2$, which is contractible. Also ${\pi }_1(V,b)=1$. All hypotheses of the Seifert-van Kampen theorem are now checked, and it gives the fundamental group of a sphere with an attached chord:where both maps from the amalgamating infinite cyclic group are trivial. A group generator follows the chord from one endpoint to the other and returns along any path on the sphere, with a connecting path from the chosen base point.