If f:X→Yf:X\to Yf:X→Y is a function, then the codomain of fff is the set YYY. The codomain is not determined solely by the rule x↦f(x)x\mapsto f(x)x↦f(x); it is specified as part of the function’s type. Many notions (notably surjectivity) depend on the codomain, not just on the actual outputs attained. Examples: If f:R→Rf:\mathbb{R}\to\mathbb{R}f:R→R is given by f(x)=x2f(x)=x^2f(x)=x2, then the codomain is R\mathbb{R}R even though f(x)≥0f(x)\ge 0f(x)≥0 always. If the same rule is viewed as f:R→[0,∞)f:\mathbb{R}\to[0,\infty)f:R→[0,∞), then the codomain is [0,∞)[0,\infty)[0,∞). The codomain of sin⁡:R→R\sin:\mathbb{R}\to\mathbb{R}sin:R→R is R\mathbb{R}R (even though the range is contained in [−1,1][-1,1][−1,1]).

Let (X,d)(X,d)(X,d) be a metric space and let K⊆XK\subseteq XK⊆X with the subspace metric. Theorem: The following are equivalent: KKK is compact . KKK is complete and totally bounded . This theorem is the fundamental metric characterization of compactness: “no missing limits” (completeness) plus “finite ε\varepsilonε-approximation at every scale” (total boundedness). Proof sketch: (⇒\Rightarrow⇒) If KKK is compact, then KKK is complete (every Cauchy sequence has a convergent subsequence , and hence the full sequence converges) and totally bounded (the balls {HAHAHUGOSHORTCODE558s7HBHB(x,ε):x∈K}\{$B$ (x,\varepsilon):x\in K\}{HAHAHUGOSHORTCODE558s7HBHB(x,ε):x∈K} form an open cover ; take a finite subcover). ...

Let (X,d)(X,d)(X,d) be a metric space and let K⊆XK\subseteq XK⊆X. The set KKK is compact if for every family of open sets {Ui}i∈I\{U_i\}_{i\in I}{Ui​}i∈I​ in XXX such that K⊆⋃i∈IUi,K\subseteq \bigcup_{i\in I} U_i,K⊆i∈I⋃​Ui​, there exist finitely many indices i1,…,iN∈Ii_1,\dots,i_N\in Ii1​,…,iN​∈I such that ...

Let K⊆RkK\subseteq \mathbb{R}^kK⊆Rk with the Euclidean metric . Proposition (equivalent characterizations): The following are equivalent: KKK is compact (every open cover has a finite subcover). KKK is sequentially compact (every sequence in KKK has a convergent subsequence with limit in KKK). KKK is closed and bounded . This packages together the key compactness theorems specialized to Euclidean spaces (Bolzano–Weierstrass + Heine–Borel + compactness–sequential compactness equivalence). ...

Compactness implies boundedness: Let (X,d)(X,d)(X,d) be a metric space and let K⊆XK\subseteq XK⊆X be compact . Then KKK is bounded : there exist x0∈Xx_0\in Xx0​∈X and R>0R>0R>0 such that K⊆B(x0,R). K\subseteq B(x_0,R). K⊆B(x0​,R). This is one of the basic “finiteness” consequences of compactness and is used to control sequences and covers. ...

Compactness implies closedness: Let (X,d)(X,d)(X,d) be a metric space and let K⊆XK\subseteq XK⊆X be compact . Then KKK is closed in XXX. This is a key structural property: in metric spaces, compact sets behave like “closed and bounded” objects, and this is one half of that picture. Proof sketch: Let (xn)(x_n)(xn​) be a sequence in KKK converging in XXX to some xxx. By sequential compactness of KKK, there is a subsequence (xnj)(x_{n_j})(xnj​​) converging to some y∈Ky\in Ky∈K. But subsequences of a convergent sequence converge to the same limit, so y=xy=xy=x. Hence x∈Kx\in Kx∈K, proving KKK is closed. ...

Compactness implies completeness: If (X,d)(X,d)(X,d) is a compact metric space , then XXX is complete . This shows compactness is a strong finiteness condition: it forces not only boundedness but also the existence of limits for all Cauchy sequences . Proof sketch (optional): Let (xn)(x_n)(xn​) be Cauchy in XXX. By compactness (sequential compactness ), it has a convergent subsequence xnk→xx_{n_k}\to xxnk​​→x. Cauchy-ness forces the full sequence to converge to the same xxx, hence (xn)→x∈X(x_n)\to x\in X(xn​)→x∈X. ...

Compactness implies total boundedness: If (X,d)(X,d)(X,d) is a compact metric space , then for every ε>0\varepsilon>0ε>0 there exist points x1,…,xN∈Xx_1,\dots,x_N\in Xx1​,…,xN​∈X such that X⊆⋃j=1NB(xj,ε).X\subseteq \bigcup_{j=1}^N B(x_j,\varepsilon).X⊆⋃j=1N​B(xj​,ε). Equivalently, XXX is totally bounded . ...

Compactness of graphs lemma: Let (X,dX)(X,d_X)(X,dX​) and (Y,dY)(Y,d_Y)(Y,dY​) be metric spaces , let K⊆XK\subseteq XK⊆X be compact , and let f:K→Yf:K\to Yf:K→Y be continuous . The graph of fff is Γf={(x,f(x))∈X×Y:x∈K}. \Gamma_f=\{(x,f(x))\in X\times Y: x\in K\}. Γf​={(x,f(x))∈X×Y:x∈K}. Then Γf\Gamma_fΓf​ is compact in the product metric space X×YX\times YX×Y. ...

Comparison Test: Let 0≤an≤bn0\le a_n\le b_n0≤an​≤bn​ for all sufficiently large nnn. If ∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞​bn​ converges , then ∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞​an​ converges. If ∑n=1∞an\sum_{n=1}^\infty a_n∑n=1∞​an​ diverges , then ∑n=1∞bn\sum_{n=1}^\infty b_n∑n=1∞​bn​ diverges. This test reduces convergence questions to bounding terms by simpler expressions. ...