hcju/proofwikiqa huggingface dataset:
Let $\epsilon \in \R_{>0}$ be a strictly positive real number. By definition, $M$ is totally bounded only if there exists a finite $\epsilon$-net for $M$. We use a Proof by Contraposition. To that end, suppose that there exists no finite $\epsilon$-net for $M$. The aim is to construct an infinite sequence $\sequence {x_n} _{n \in \N}$ in $A$ that has no convergent subsequence. Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that: :$\forall m, n \in \set {0, 1, \ldots, r} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$ That is, any two distinct elements of $\set {x_0, x_1, \ldots, x_r}$ are at least $\epsilon$ apart. By hypothesis, there exists no finite $\epsilon$-net for $M$. Specifically, $\set {x_0, x_1, \ldots, x_r}$ is therefore not an $\epsilon$-net for $M$. So, by definition of a $\epsilon$-net: :$\displaystyle A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$ where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$ in $M$. Thus there must exist $x_{r+1} \in A$ such that: :$\displaystyle x_{r+1} \notin \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$ That is: :$\exists x_{r+1} \in A: \forall m, n \in \set {0, 1, \ldots, r, r + 1} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$ Thus, by induction, the infinite sequence $\sequence {x_n}$ in $A$ has been constructed such that: :$\forall m, n \in \N: m \ne n \implies \map d {x_m, x_n} \ge \epsilon$ Thus $\sequence {x_n}$ has no Cauchy subsequence. Since a convergent sequence is Cauchy, it has no convergent subsequence either.  Thus, by definition, $M$ is not sequentially compact. From Rule of Transposition, if $M$ is sequentially compact, then there exists no finite $\epsilon$-net for $M$. So if $M$ is sequentially compact, it is certainly not totally bounded. {{qed}}
For any open set $U \subseteq S$ such that $x \in U$, we must have $U = S$. Hence: : $\forall n \ge 1: x_n \in U$ The result follows from the definition of a convergent sequence in a topological space. {{qed}}
Let $M = \left({A, d}\right)$ be a totally bounded metric space. By the definition of total boundedness, we can use the axiom of countable choice to construct a sequence $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that: :For all natural numbers $n \ge 1$, $F_n$ is a finite $\left({1/n}\right)$-net for $M$. Let $\displaystyle S = \bigcup_{n \mathop \ge 1} F_n$. From Countable Union of Countable Sets is Countable, it follows that $S$ is countable. It suffices to prove that $S$ is everywhere dense in $M$. Let $S^-$ denote the closure of $S$. Let $x \in X$. Let $U \subseteq X$ be open in $M$ such that $x \in U$. By definition, there exists a strictly positive real number $\epsilon$ such that $B_{\epsilon} \left({x}\right) \subseteq U$. That is, the open $\epsilon$-ball of $x$ in $M$ is contained in $U$. By the Archimedean Principle, there exists a natural number $n > \dfrac 1 \epsilon$. That is, $\dfrac 1 n < \epsilon$, and so $B_{1 / n} \left({x}\right) \subseteq B_{\epsilon} \left({x}\right)$. Since $\subseteq$ is a transitive relation, we have $B_{1/n} \left({x}\right) \subseteq U$. By the definition of a net, there exists a $y \in F_n$ such that $x \in B_{1/n} \left({y}\right)$. By axiom $\left({M3}\right)$ for a metric, it follows from the definition of an open ball that $y \in B_{1/n} \left({x}\right)$. Since $y \in S \cap U$, it follows that $x$ is an adherent point of $S$. By definition of adherent point, we have $x \in S^-$. That is, $X \subseteq S^-$, and so $S$ is everywhere dense in $M$. {{qed}} {{ACC}} Category:Separable Spaces Category:Metric Spaces Category:Totally Bounded Metric Spaces ejynvni81404dwnx6gyadutj5l949qs
From the definition of convergent sequence, we have that $\sequence {x_n}$ converges to at least one limit. {{AimForCont}} $\ds \lim_{n \mathop \to \infty} x_n = l$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = m$ such that $l \ne m$. As $T$ is Hausdorff, $\exists U \in \tau: l \in U$ and $\exists V \in \tau: m \in V$ such that $U \cap V = \O$. Then, from the definition of convergent sequence: {{begin-eqn}} {{eqn | ll= \exists N_U \in \R:       | l = n > N_U       | o = \implies       | r = x_n \in U }} {{eqn | ll= \exists N_V \in \R:       | l = n > N_V       | o = \implies       | r = x_n \in V }} {{end-eqn}} Taking $N = \max \set {N_U, N_V}$ we then have: :$\exists N \in \R: n > N \implies x_n \in U, x_n \in V$ But $U \cap V = \O$. From that contradiction we can see that there can be no such two distinct $l$ and $m$. Hence the result. {{qed}}
Let $x, y \in A: x \ne y$. Then from Distinct Points in Metric Space have Disjoint Open Balls, there exist open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} Y$ which are disjoint open sets containing $x$ and $y$ respectively. Hence the result by the definition of Hausdorff space. {{qed}}
{{AimForCont}} that $\sequence {s_n}$ converges to $l$ and also to $m$. That is, suppose that: :$\displaystyle \lim_{n \mathop \to \infty} x_n = l$ and: :$\displaystyle \lim_{n \mathop \to \infty} x_n = m$ {{WLOG}}, assume that $l \ne m$. Let: :$\epsilon = \dfrac {\size {l - m} } 2$ As $l \ne m$, it follows that $\epsilon > 0$. As $\sequence {s_n} \to l$: :$\exists N_1 \in \N: \forall n \in \N: n > N_1: \size {s_n - l} < \epsilon$ Similarly, since $\sequence {s_n} \to m$: :$\exists N_2 \in \N: \forall n \in \N: n > N_2: \size {s_n - m} < \epsilon$ Now set $N = \max \set {N_1, N_2}$. We have: {{begin-eqn}} {{eqn | l = \size {l - m}       | r = \size {l - s_N + s_N - m}       | c =  }} {{eqn | o = \le       | r = \size {l - s_N} + \size {s_N - m}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = <       | r = 2 \epsilon       | c =  }} {{eqn | r = \size {l - m}       | c =  }} {{end-eqn}} This constitutes a contradiction. It follows from Proof by Contradiction that $l = m$. {{qed}}
We have that the real number line is a metric space The result then follows from Convergent Sequence in Metric Space has Unique Limit. {{qed}}
Let $x, y \in A: x \ne y$. Then $d \left({x, y}\right) > 0$. Put $\epsilon = \dfrac {d \left({x, y}\right)} 2$. Let $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ be the open $\epsilon$-balls of $x$ and $y$ respectively. Suppose $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ are not disjoint. Then $\exists z \in M$ such that $z \in B_\epsilon \left({x}\right)$ and $z \in B_\epsilon \left({y}\right)$. Then $d \left({x, z}\right) < \epsilon$ and $d \left({z, y}\right) < \epsilon$. Hence $d \left({x, z}\right) + d \left({z, y}\right) < 2 \epsilon = d \left({x, y}\right)$. This contradicts the definition of a metric, so there can be no such $z$. Hence the open balls must be disjoint. {{qed}} Category:Open Balls qfipa27r1dmttasxcnjn6xadd0fmje2
Let $M = \left({X, d}\right)$ be a sequentially compact metric space. Take any open cover $C$ of $M$. We need to find a countable subset of $C$ which still covers $X$. We have that a Sequentially Compact Metric Space is Second-Countable. Thus, by definition, the topology of $M$ has a countable basis. Let $\mathcal B$ be a countable basis for the topology induced by $d$ of $M$. Let $x \in X$. As $C$ covers $X$: :$\exists U_x \in C: x \in U_x$ As $\mathcal B$ is a basis: :$\exists B_x \in \mathcal B: x \in B_x \subseteq U_x$ Thus: : $(1): \quad \forall x \in X: \exists B_x \in \mathcal B: x \in B_x$ Consider the set $\Sigma := \left\{{B_x: x \in X}\right\}$. $\Sigma$ is a subset of $\mathcal B$. Hence $\Sigma$ is countable. As $\Sigma$ contains every $x \in X$ from $(1)$, $\Sigma$ covers $X$. By construction of $\Sigma$, every open set in $\Sigma$ is contained in some $U \in C$. For each open set $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$. {{explain|Does this need AoC or ACC?}} Let $\mathcal U$ be the set defined as: :$\mathcal U = \left\{ {U_B: B \in \Sigma}\right\}$ As $\mathcal U$ does not have more sets than $\Sigma$, $\mathcal U$ is countable. We have that: : $\forall B \in \Sigma: B \subseteq U_B$ Thus it follows that $\mathcal U$ covers $X$. As: :$\forall U_B \in \mathcal U: U_B \in C$ $U$ is a countable subcover of $C$. Thus a countable subcover has been obtained from $C$. As $C$ is arbitrary, it follows that $M$ is a Lindelöf space. {{qed}}
Let $\struct {A, d}$ be a sequentially compact metric space. From Sequentially Compact Metric Space is Lindelöf, every open cover of $A$ has a countable subcover. Let $C$ be an open cover $A$. Extract from it a countable subcover $\set {U_1, U_2, \ldots}$. {{AimForCont}} there exists no finite subcover of $C$. Then, for all $n \in \N_{\ge 1}$, the set $\set {U_1, \ldots, U_n}$ does ''not'' cover $A$. Hence it is possible to choose $x_n \in A$ such that: :$x_n \notin U_1 \cup \cdots \cup U_n$. Thus we construct an infinite sequence $\sequence {x_n}_{n \mathop \ge 1}$ of points of $A$. By assumption $A$ is sequentially compact. Thus $\sequence {x_n}_{n \mathop \ge 1}$ has a subsequence which converges to some $x \in A$. But because the $U_i: i \ge 1$ forms a cover for $A$, there exists some $U_m$ such that $x \in U_m$. Hence by one of the characterizations of convergence, there is an infinite number of terms in the sequence $\sequence {x_i}$ which are contained in $U_m$. {{explain|Determine which of those characterizations of convergence.}} But from the method of construction of $\sequence {x_i}$, each $U_n$ can contain only points $x_i$ with $i < n$. That is, each $U_n$ can only contain a finite number of the terms of $\sequence {x_i}$. This is a contradiction. Thus the supposition that there exists no finite subcover of $C$ was false. Hence the result. {{qed}}
Let $M = \left({A, d}\right)$ be a sequentially compact metric space. Let $\mathcal U$ be any open cover of $M$. By Lebesgue's Number Lemma, there exists a Lebesgue number for $\mathcal U$. By Sequentially Compact Metric Space is Totally Bounded, there exists $\left\{{x_1, x_2, \ldots, x_n}\right\}$ which is a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number. Let $B_\epsilon \left({x_i}\right)$ be the open $\epsilon$-ball of $x_i$. By definition of Lebesgue number, $B_\epsilon \left({x_i}\right)$ is contained in some $U_i \in \mathcal U$. Since: : $\displaystyle M \subseteq \bigcup_{i \mathop = 1}^n B_\epsilon \left({x_i}\right) \subseteq \bigcup_{i \mathop = 1}^n U_i$ we have a finite subcover $\left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal{U}$ for $M$. Hence the result. {{qed}}
Follows directly from: : Sequentially Compact Space is Countably Compact : Countably Compact Metric Space is Compact {{qed}}
Let $a \in M$. Let $n \in \N_{>0}$. Let $B_n \left({a}\right)$ be the open $n$-ball of $a$. Then $\displaystyle C \subseteq \bigcup_{n \mathop = 1}^\infty B_n \left({a}\right)$ because $\forall x \in C: d \left({x, a}\right) < n$ for some $n \in \N$. Thus the collection $\left\{{B_n \left({a}\right): n \in \N}\right\}$ forms an open cover of $C$. Because $C$ is compact, it has a finite subcover, say: $\left\{{B_{n_1} \left({a}\right), B_{n_2} \left({a}\right), \ldots, B_{n_r} \left({a}\right)}\right\}$. Let $n = \max \left\{{n_1, n_2, \ldots, n_r}\right\}$. Then: : $\displaystyle C \subseteq \bigcup_{n \mathop = 1}^r B_{n_r} \left({a}\right) = B_n \left({a}\right)$ The result follows by definition of bounded. {{qed}}
For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$. Consider an arbitrary point $x \in C^{\complement}$. Define the set: :$\displaystyle \mathcal O = \left\{{V \in \tau: \exists U \in \tau: x \in U \subseteq V^{\complement}}\right\}$ By Empty Intersection iff Subset of Complement, we have that: :$U \subseteq V^{\complement} \iff U \cap V = \varnothing$ Hence, by the definition of a Hausdorff space, it follows that $\mathcal O$ is an open cover for $C$. By the definition of a compact subspace, there exists a finite subcover $\mathcal F$ of $\mathcal O$ for $C$. By the Principle of Finite Choice, there exists an $\mathcal F$-indexed family $\left\langle{U_V}\right\rangle_{V \in \mathcal F}$ of elements of $\tau$ such that: :$\forall V \in \mathcal F: x \in U_V \subseteq V^{\complement}$ Define: :$\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U_V$ By General Intersection Property of Topological Space, it follows that $U \in \tau$. Clearly, $x \in U$. We have that: {{begin-eqn}} {{eqn | l = U       | o = \subseteq       | r = \bigcap_{V \mathop \in \mathcal F} \left({ V^{\complement} }\right)       | c = Set Intersection Preserves Subsets }} {{eqn | r = \left({\bigcup \mathcal F}\right)^{\complement}       | c = De Morgan's laws }} {{eqn | o = \subseteq       | r = C^{\complement}       | c = {{Defof|Cover of Set}} and Relative Complement inverts Subsets }} {{end-eqn}} From Subset Relation is Transitive, we have that $U \subseteq C^{\complement}$. Hence $C^{\complement}$ is a neighborhood of $x$. From Set is Open iff Neighborhood of all its Points, we have that $C^{\complement} \in \tau$. That is, $C$ is closed in $H$. {{qed}}
From Subspace of Hausdorff Space is Hausdorff, a subspace of a Hausdorff space is itself Hausdorff. Let $a \in A \setminus C$. We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$. For any single point $x \in C$, the Hausdorff condition ensures the existence of disjoint open set $\map U x$ and $\map V x$ containing $a$ and $x$ respectively. Suppose there were only a finite number of points $x_1, x_2, \ldots, x_r$ in $C$. Then we could take $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$. Now suppose $C$ is not finite. The set $\set {\map V x: x \in C}$ is an open cover for $C$. As $C$ is compact, it has a finite subcover, say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$. Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$. Then $U_a$ is open because it is a finite intersection of open sets. Also, $a \in U_a$ because $a \in \map U {x_i}$ for each $i = 1, 2, \ldots, r$. Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in \map U {x_i}$. Because $\displaystyle C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$: :$b \notin \map V {x_i}$, so $b \notin C$ Thus: :$U_a \subseteq A \setminus C$. Then: :$\displaystyle A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$ So $A \setminus C$ is open. It follows that $C$ is closed. {{qed}}
Suppose $\UU$ is an open cover of $f \sqbrk {T_1}$ by sets open in $T_2$. Because $f$ is continuous, it follows that $f^{-1} \sqbrk U$ is open in $T_1$ for all $U \in \UU$. The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an open cover of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \UU$. Because $T_1$ is compact, it has a finite subcover $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$. It follows that $\set {U_1, U_2, \ldots, U_r}$ is a finite subcover of $f \sqbrk {T_1}$. {{qed}}
Let $T$ be a compact space. Let $C$ be a closed subspace of $T$. Let $\UU$ be an open cover of $C$. Since $C$ is closed, it follows by definition of closed that $T \setminus C$ is open in $T$. So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an open cover of $T$. As $T$ is compact, there is a finite subcover of $\UU \cup \set {T \setminus C}$, say $\VV = \set {U_1, U_2, \ldots, U_r}$. This covers $C$ by the fact that it covers $T$. If $T \setminus C$ is an element of $\VV$, then it can be removed from $\VV$ and the rest of $\VV$ still covers $C$. Thus we have a finite subcover of $\UU$ which covers $C$, and hence $C$ is compact. {{qed}}
Let $I$ be one of the intervals as specified in the exposition. Note that: :$(1): \quad$ By Condition for Point being in Closure, $x \in I^-$ {{iff}} every open set in $\R$ containing $x$ contains a point in $I$. :$(2): \quad$ From Union of Open Sets of Metric Space is Open, every open set in $\R$ is a union of open intervals. Thus we also have that $x \in I^-$ {{iff}} every open interval containing $x$ also contains a point in $I$. This equivalence will be made use of throughout. === Lemma: $x \in \left[{a \,.\,.\, b}\right] \implies x \in I^-$ === Let $x \in \left[{a \,.\,.\, b}\right]$. Let $\left({c \,.\,.\, d}\right)$ be an open interval in $\R$ such that $x \in \left({c \,.\,.\, d}\right)$. We must show that $\left({c \,.\,.\, d}\right)$ contains a point in $I$. One of the following three possibilities holds: : $a < x < b$ : $x = a$ : $x = b$ ==== Case: $a < x < b$ ==== In this case, $x \in I$ and $x \in \left({c \,.\,.\, d}\right)$. Therefore $\left({c \,.\,.\, d}\right)$ contains a point in $I$. {{qed|lemma}} ==== Case: $x = a$ ==== If $I$ contains $a$, then this means $x \in I$, and the proof is complete. So, assume that $a \notin I$. Since $I$ is nonempty but does not contain $a$, we must have $a < b$. Let $r$ be the minimum of $d$ and $b$, so that $r \le d$ and $r \le b$. Since $a = x < d$ by choice of $d$ and since $a < b$ by assumption, we must have $a < r$. Thus, by Real Numbers are Close Packed, there exists some $s \in \R$ such that $a < s < r$. To summarize, we have $c < x = a < s < r$, where $r \le d$ and $r \le b$. This means that $s$ satisfies both $c < s < d$ and $a < s < b$. Hence, $s$ is a point in $\left({c \,.\,.\, d}\right)$ which is also in $I$. The existence of such a point is what we wanted to show. {{qed|lemma}} ==== Case: $x = b$ ==== This case is analogous to case when $x = a$. Here we instead let $l$ be the maximum of $c$ and $a$, and select an $s$ such that $l < s < x = b < d$, where $c \le l$ and $a \le l$. {{qed|lemma}} === Lemma: $x \notin \left[{a \,.\,.\, b}\right] \implies x \notin I^-$  === Suppose $x \notin \left[{a \,.\,.\, b}\right]$. We must find an open interval containing $x$ which does not contain a point in $I$. There are two possibilities: : $x < a$ or: : $x > b$ ==== Case: $x < a$ ==== By Real Numbers are Close Packed, there exists $r \in \R$ such that $x < r < a$. Thus $\left({x-1 \,.\,.\, r}\right)$ is an open interval, all of whose elements are less than $a$, and hence not in $I$. ==== Case: $x > b$ ==== This is similarly to the case when $x < a$. Here instead we pick $r$ such that $b < r < x$, and consider the interval $\left({r \,.\,.\, x+1}\right)$. {{qed|lemma}} By the two lemmas proven above: : $\left[{a \,.\,.\, b}\right] = I^-$ {{qed}}
There are four cases to cover: :$(1): \quad$ Let $I = \left({a \,.\,.\, b}\right)$. From Closure of Open Real Interval is Closed Real Interval: :$I^- = \left[{a \,.\,.\, b}\right]$ {{qed|lemma}} :$(2): \quad$ Let $I = \left[{a \,.\,.\, b}\right)$. From Closure of Half-Open Real Interval is Closed Real Interval: :$I^- = \left[{a \,.\,.\, b}\right]$ {{qed|lemma}} :$(3): \quad$ Let $I = \left({a \,.\,.\, b}\right]$. From Closure of Half-Open Real Interval is Closed Real Interval: :$I^- = \left[{a \,.\,.\, b}\right]$ {{qed|lemma}} :$(4): \quad$ Let $I = \left[{a \,.\,.\, b}\right]$. From Closed Real Interval is Closed in Real Number Line: :$I$ is closed in $\R$. From Set is Closed iff Equals Topological Closure: :$I^- = \left[{a \,.\,.\, b}\right]$ {{qed|lemma}} Thus all cases are covered. The result follows by Proof by Cases. {{qed}}
Let $T_1 \times T_2$ be compact. From Projection from Product Topology is Continuous, both $T_1$ and $T_2$ are the continuous images of $T_1 \times T_2$ under the projections $\pr_1$ and $\pr_2$ respectively. So both $T_1$ and $T_2$ are compact by Continuous Image of Compact Space is Compact. Now let $T_1$ and $T_2$ be compact. Let $\WW$ be an open cover for $T_1 \times T_2$. Let $A \subseteq T_1$ be defined as $\text{good}$ (for $\WW$) if $A \times T_2$ is covered by a finite subset of $\WW$. We want to prove that $T_1$ is $\text{good}$. We will do this in stages, because it's complicated. === Step 1 === Suppose $A_1, A_2, \ldots, A_r$ are all $\text{good}$. Then so is $\displaystyle A = \bigcup_{i \mathop = 1}^r A_i$. For any given $i=1, 2, \ldots, r$ we have that $A_i \times T_2$ is covered by a finite subset of $\WW$, say $\WW_i \subseteq \WW$. Hence $\displaystyle A \times T_2 = \bigcup_{i \mathop = 1}^r \paren{A_i \times T_2}$ is covered by the finite subset $\displaystyle \bigcup_{i \mathop = 1}^r \WW_i$ of $\WW$. === Step 2 === Now we show that $T_1$ is '''locally''' $\text{good}$, in the sense that $\forall x \in T_1$, there is an open set $\map U x$ such that $x \in \map U x$ and $\map U x$ is $\text{good}$. Consider a fixed $x \in T_1$. For each $y \in T_2$, we have that $\tuple{x, y} \in \map W y$ for some $\map W y \in \WW$, since $\WW$ covers $T_1$ and $T_2$. By the definition of the product topology, $\exists \map U y, \map V y$ open in $T_1$ and $T_2$ respectively such that $\tuple{x, y} \in \map U y \times \map V y \subseteq \map W y$. The set $\set{\map V y : y \in T_2}$ is an open cover for $T_2$. As $T_2$ is compact, there is a finite subcover of $\map V y$, say $\set{\map V {y_1}, \map V {y_2}, \ldots, \map V {y_r}}$. Let $\map U x = \map U {y_1} \cap \map U {y_2} \cap \cdots \cap \map U {y_r}$. For each $i = 1, 2, \ldots, r$, we have $\map U x \times \map V {y_i} \subseteq \map U {y_i} \times \map V {y_i} \subseteq \map W {y_i}$. Hence $\displaystyle \map U x \times T_2 = \map U x \times \bigcup_{i \mathop = 1}^r \map V {y_i} \subseteq \map W {y_i}$. So $\map U x$ is $\text{good}$. And we have that $x \in \map U x$ and $\map U x$ is open in $T_1$ as required. === Step 3 === Now we pass from local to global. For each $x \in T_1$, let $\map U x$ be a $\text{good}$ open set containing $x$, by step 2. Then $\set{\map U x: x \in T_1}$ is an open cover for $T_1$. Because $T_1$ is compact, $\map U x$ has a finite subcover, say $\set{\map U {x_1}, \map U {x_2}, \ldots, \map U {x_m}}$. Since each $\map U {x_i}$ is $\text{good}$, then so is $\displaystyle \bigcup_{i \mathop = 1}^m \map U {x_i}$ by step 1. But $\displaystyle \bigcup_{i \mathop = 1}^m \map U {x_i} = T_1$, so $T_1$ is $\text{good}$, as required. {{qed}}
Let $\epsilon \in \R_{>0}$ be a strictly positive real number. For all $x \in A_1$, define: :$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$ Define: :$\mathcal C = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map \Delta x}$ where $B_{\delta} \left({x}\right)$ denotes the open $\delta$-ball of $x$ in $M_1$. From the definition of continuity, it follows that $\mathcal C$ is a cover for $A_1$. From Open Ball of Metric Space is Open Set, it therefore follows that $\mathcal C$ is an open cover for $A_1$. By the definition of a compact metric space, there exists a finite subcover $\set {\map {B_{\delta_1} } {x_1}, \map {B_{\delta_2} } {x_2}, \ldots, \map {B_{\delta_n} } {x_n} }$ of $\mathcal C$ for $A_1$. Define: :$\delta = \min \set {\delta_1, \delta_2, \ldots, \delta_n}$ Let $x, y \in A_1$ satisfy $\map {d_1} {x, y} < \delta$. By the definition of a cover, there exists a $k \in \set{1, 2, \ldots, n}$ such that $\map {d_1} {x, x_k} < \delta_k$. Then: {{begin-eqn}} {{eqn | l = \map {d_1} {y, x_k}       | o = \le       | r = \map {d_1} {y, x} + \map {d_1} {x, x_k}       | c = Triangle Inequality }} {{eqn | o = <       | r = \delta + \delta_k       | c = Metric Space Axiom $\paren {M3}$ }} {{eqn | o = \le       | r = 2 \delta_k }} {{end-eqn}} By the definition of $\Delta \left({x_k}\right)$, it follows that: :$\map {d_2} {\map f x, \map f {x_k} } < \dfrac \epsilon 2$ :$\map {d_2} {\map f y, \map f {x_k} } < \dfrac \epsilon 2$ Hence: {{begin-eqn}} {{eqn | l = \map {d_2} {\map f x, \map f y}       | o = \le       | r = \map {d_2} {\map f x, \map f {x_k} } + \map {d_2} {\map f {x_k}, \map f y}       | c = Triangle Inequality }} {{eqn | o = <       | r = \frac \epsilon 2 + \frac \epsilon 2       | c = Metric Space Axiom $\paren{M3}$ }} {{eqn|r = \epsilon }} {{end-eqn}} The result follows from the definition of uniform continuity. {{qed}}