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Good evening, I'm trying to prove that

Let $X$ be a metric space. If every sequence in $X$ has a cluster point in $X$ $\implies$ every open cover of $X$ has a finite subcover.

Actually, I'm trying to prove

Let $X$ be a metric space. Prove that the following statements are equivalent.

(i) Every open cover of $X$ has a finite subcover.

(ii) $X$ is totally bounded and complete.

(iii) Every sequence in $X$ has a cluster point in $X$.

I've just proved (i) implies (ii) and (ii) implies (iii). The only remaining is (iii) implies (i).

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!


My attempt:

Let $d$ be the metric on $X$.

First, we prove that $X$ is totally bounded. Suppose the contrary that $X$ is not totally bounded.

  • Then there exists $r>0$ such that $X \not \subseteq \bigcup_{k=0}^{m} \mathbb{B}\left(x_{k}, r\right)$ for any finite set $\{x_{0}, \ldots, x_{m}\} \subseteq K .$ In particular, there exists $x_{0} \in K$ such that $X \not \subseteq \mathbb{B} (x_{0}, r) .$ Thus there exists $x_{1} \in \left(\mathbb{B}(x_{0}, r)\right)^c$. Since $X \not \subseteq \left (\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)$, there exists $x_{2} \in \left(\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)^c$. Continuing in this way and with Axiom of Countable Choice, there is a sequence $(x_n)$ in $X$ such that $x_{n+1} \in \left(\bigcup_{k=0}^n \mathbb{B}(x_{k}, r)\right)^c$. It follows that $x_{n+1} \notin \mathbb{B}(x_{k}, r)$ and thus $d(x_k, x_{n+1}) \ge r$ for all $k \le n$.

  • By our hypothesis, $(x_n)$ has a cluster point, i.e. there exists a subsequence $(x_{\psi(n)})$ of $(x_n)$ such that $x_{\psi(n)} \to \bar x$ as $n \to \infty$. It follows from $x_{\psi(n)} \to \bar x$ that there is $N \in \mathbb N$ such that $d(x_{\psi(n)},\bar x) < r/2$ for all $n \ge N$. Hence $d(x_{\psi(N)},\bar x) < r/2$ and $d(x_{\psi(N+1)},\bar x) < r/2$. It follows that $d(x_{\psi(N)}, x_{\psi(N+1)}) \le d(x_{\psi(N)},\bar x) + d(x_{\psi(N+1)},\bar x) < r/2 +r/2 = r$. This contradicts our construction of $(x_n)$. Hence $X$ is totally bounded.

Next, we prove that every open cover $\{O_i \mid i \in I \}$ of $X$ has a countable subcover.

  • Because $X$ is totally bounded, for each $n \ge 1$ there are finitely many $x^i_n$ such that $X = \bigcup_{i=0}^{k_n} \mathbb B (x_n^i , 1/n)$. Let $A = \bigcup_{n=1}^\infty \{x_n^0, \ldots, x_n^{k_n}\}$. Because $A$ is countable union of countable sets, $A$ is countable.

  • We define a mapping $f:A \times \mathbb Q_+ \to I$ by corresponding (with help from Axiom of Choice) $(a,r) \in A \times \mathbb Q_+$ with an $i$ such that $\mathbb B(a,r) \subseteq O_i$ if such $i$ exists, otherwise $f(a,r) =O_{i_0}$ for some $i_0 \in I$. Let $J=f[I]$. Because $A,\mathbb Q_+$ are countable, $A \times \mathbb Q_+$ is countable and so is $J$.

  • For $x\in X$, there exists some $j \in I$ such that $x \in O_j$. Because $O_j$ is open, there is $r>0$ such that $\mathbb B(x,r) \subseteq O_j$. Then we choose some $a \in A$ such that $d(a,x) < r/2$ and some $r' \in \mathbb Q$ such that $d(a,x) <r' < r/2$. It follows that $x \in \mathbb B(a,r') \subseteq \mathbb B(x,r) \subseteq O_j$ by triangle inequality. By the construction of $f$, $x \in O_{f(a,r')}$. As such, $\{O_i \mid i \in J\}$ is a countable subcover of $X$.

Finally, we prove that $X$ is compact. Let $\{O_i \mid i \in I \}$ be an open cover of $X$.

  • We've just proved that there exists a countable subcover $\{O_k \mid k \in \mathbb N\}$ of $\{O_i \mid i \in I \}$. Assume the contrary that $\{O_i \mid i \in I \}$ has no finite subcover. Then $X \not \subseteq \bigcup_{k =0}^n O_k$ for any $k \in \mathbb N$. As such, $X_n:=\bigcap_{k=0}^n O^c_{k} = \left (\bigcup_{k=0}^n O_{k} \right)^c \neq \emptyset$ for all $n \in \mathbb N$. On the other hand, $(O_k)_{k \in \mathbb N}$ is a subcover of $X$, so $\bigcup_{k=0}^\infty O_k =X$ or equivalently $\bigcap_{k=0}^\infty X_k = \bigcap_{k=0}^\infty O^c_{k} = \emptyset$.

  • By Axiom of Countable Choice, we define the sequence $(x_n)$ in $X$ recursively by $x_{n+1} \in X_n:= \bigcap_{k=0}^{n+1} O^c_{k}$ for all $n \in \mathbb N$. It follows that $X_{n+1} \subseteq X_n$ and that $X_n$ is closed in $X$ for all $n$. By hypothesis, $(x_n)$ has a cluster point $\bar x \in X$, i.e. there is a subsequence $(x_{\phi(n)})$ of $(x_n)$ such that $x_{\phi(n)} \to \bar x \in X$. It follows our construction of $(x_n)$ that $x_{\phi(n)} \in X_N$ for all $n \ge N$. Moreover, $X_N$ is closed, so $\bar x \in X_N$. Because this is true for all $N$, we have $\bar x \in \bigcap_{k=0}^\infty X_k = \emptyset$. This is a contradiction. Hence $\{O_i \mid i \in I \}$ has a finite subcover.

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  • $\begingroup$ What does your notation $f[I]$ mean? $I$ is the codomain of $f$. $\endgroup$ – Paul Sinclair Oct 29 '19 at 23:54
  • $\begingroup$ "$f(a,r) =O_{i_0}$" doesn't make sense, $I$ is the codomain of $f$, but it is $i_0 \in I$, not $O_{i_0}$. Did you mean $f(a,r) = i_0$? $\endgroup$ – Paul Sinclair Oct 29 '19 at 23:58
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    $\begingroup$ Your final bullet is more convoluted than it needs to be. That $X_{n+1} \subseteq X_n$ and $X_n$ is closed belong in the previous bullet. They do not follow from what immediately preceeds them now. $x_{\phi(n)} \in X_{\phi(n)}$ by construction, and in $X_N, N \le \phi(n)$ by the nesting property. That it works for $n$ instead of $\phi(n)$ is only true because of the nature of subsequence indexing, and is not needed here. However, the logic appears to be correct, assuming you really meant $J = f(A\times \Bbb Q_+)$. $\endgroup$ – Paul Sinclair Oct 30 '19 at 0:29
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    $\begingroup$ Hi @PaulSinclair, I'm really grateful for your time reading my long proof: (1) Yeap, they are typos, I meant $J=f[A \times \mathbb Q_+]$ and $f(a,r) =i_0$ for some fixed $i_0 \in I$. (2) If $n \ge N$ then $\phi(n) \ge n \ge N$. Thus $x_{\phi(n)} \in X_N$ for all $n \ge N$. $\endgroup$ – Akira Oct 30 '19 at 6:47
  • $\begingroup$ I would like to thank you you again for your kindness @PaulSinclair :))) My proof is really long :( $\endgroup$ – Akira Oct 30 '19 at 6:48
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Here is my attempt on proving the equivalence of these three statements. It would be great if someone helps me verify the other two attempts. Thank you so much @Paul Sinclair for your very kindness!


Theorem: Let $X$ be a metric space. Prove that the following statements are equivalent.

(i) Every open cover of $X$ has a finite subcover.

(ii) $X$ is totally bounded and complete.

(iii) Every sequence in $X$ has a cluster point in $X$.

My Attempt:

  1. Every open cover of $X$ has a finite subcover $\implies$ $X$ is totally bounded and complete

For $\epsilon > 0$, $\{\mathbb B(x,\epsilon) \mid x \in X\}$ is an open cover of $X$. Then there is a finite set $I \subseteq X$ such that $\cup_{x \in I} \mathbb B(x,\epsilon)$ covers $X$. Hence $X$ is totally bounded.

Let $(x_n)$ be a Cauchy sequence in $X$. Assume the contrary that $(x_n)$ does not converge to any point in $X$. Then $(x_n)$ has no cluster point in $X$. Thus, for any $x \in X$, there is a neighborhood $\mathcal U_x$ of $x$ such that $\mathcal U_x$ contains at most finitely many terms of $(x_n)$. Because $\{\mathcal U_x \mid x \in X\}$ is an open cover of $X$, there is a finite set $I \subseteq X$ such that $\bigcup_{x \in I} \mathcal U_x$ covers $X$. On the other hand, $\cup_{x \in I} \mathcal U_x$ contains at most finitely many terms of $(x_n)$. This contradiction shows that $(x_n)$ converges to some point in $X$, so $X$ is complete.

  1. $X$ is totally bounded and complete $\implies$ Every sequence in $X$ has a cluster point in $X$

Let $(x_n)$ be a sequence in $X$. We define the sequences $(y_k),(I_k)$ and a mapping $\varphi:\mathbb N_+ \to \mathbb N$ recursively as follows:

The case $k=1$:

  • Because $X$ is totally bounded, $X$ is covered by finitely many balls $B_i=\mathbb B(y_1^i,1)$ where $y_1^i \in X$ for all $i= \overline{1,n_1}$.

  • Then there exists $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i$. If not, for all $i= \overline{1,n_1}$, $B_i$ contains at most finitely many terms of $(x_n)$, and so does $\bigcup_{i=1}^{n_1} B_i$. This contradicts the fact that $\bigcup_{i=1}^{n_1} B_i$ covers $X$.

  • Let $y_{1} = y^{n_0}_{1}$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i$. Let $I_1 = \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1)\}$ and $\varphi (1) = \min \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1)\}$.

The case $k=2$:

  • Because $X$ is totally bounded, $X$ is covered by finitely many balls $B_i=\mathbb B(y_2^i,1/2)$ where $y_2^i \in X$ for all $i= \overline{1,n_2}$.

  • Then there exists $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \cap \mathbb B(y_1,1)$. If not, for all $i= \overline{1,n_2}$, $B_i \bigcap \mathbb B(y_1,1)$ contains at most finitely many terms of $(x_n)$, and so does $\bigcup_{i=1}^{n_1}\left [B_i \bigcap \mathbb B(y_1,1)\right] = \left(\bigcup_{i=1}^{n_1} B_i\right) \bigcap \mathbb B(y_1,1) = \mathbb B(y_1,1)$. This contradicts the construction of $y_1$.

  • Let $y_2 = y^{n_0}_2$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \cap \mathbb B(y_1,1)$. Let $I_2 = \{n \in \mathbb N \mid x_n \in \mathbb B(y_1,1) \bigcap \mathbb B(y_2,1/2)\}$ and $\varphi (2) = \min (I_{2} \setminus \{n \in \mathbb N \mid 1 \le n \le \varphi (1)\})$.

The inductive case:

  • We have $X$ is covered by finitely many balls $B_i :=\mathbb B(y_{k+1}^i, 1/(k+1))$ where $y_{k+1}^i \in X$ for all $i= \overline{1,n_{k+1}}$.

  • There exists $i$ such that $B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$ contains infinitely many terms of $(x_n)$. If not, $\mathbb B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$ contains at most finitely many terms of $(x_n)$ for all $i= \overline{1,n_{k+1}}$, and so does $\bigcup_{i=1}^{n_{k+1}} \left [B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right) \right] = \left(\bigcup_{i=1}^{n_{k+1}} B_i \right) \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right) =$ $Y \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)= \bigcap_{n=1}^k \mathbb B(y_n, 1/n)$. This contradicts the definition of $y_k$.

  • Let $y_{k+1} = y^{n_0}_{k+1}$ where $n_0$ is the least $i$ such that infinitely many terms of $(x_n)$ belongs to $B_i \bigcap \left (\bigcap_{n=1}^k \mathbb B(y_n, 1/n) \right)$. Let $I_{k+1} = \{m \in \mathbb N \mid x_m \in \bigcap_{n=1}^{k+1} \mathbb B(y_n, 1/n)\}$ and $\varphi (k+1) = \min (I_{k+1} \setminus \{n \in \mathbb N \mid 1 \le n \le \varphi (k)\})$.

By construction, we have

  • $\varphi$ is strictly increasing and thus $(x_{\varphi (n)})$ is a subsequence of $(x_n)$.

  • $x_{\varphi(n)} \in \mathbb B(y_k, 1/k)$ for all $k = \overline{1,n}$.

Next we prove that $(x_{\varphi (n)})$ is a Cauchy sequence. Given $\epsilon >0$, there is $N \in \mathbb N$ such that $1/N < \epsilon/2$. For all $n \ge N$, we have $x_{\varphi(n)} \in \mathbb B(y_N, 1/N)$ and $x_{\varphi (N)} \in \mathbb B(y_N, 1/N)$. Thus $\| x_{\varphi (n)} - y_N\| < 1/N$ and $\| y_N - x_{\varphi (N)}\| <1/N$. As such, $\| x_{\varphi (n)} - x_{\varphi (N)}\| \le \| x_{\varphi (n)} - y_N\| + \| y_N - x_{\varphi (N)}\|<$ $1/N +1/N < \epsilon$ for all $n > N$.

Because $X$ is complete, $(x_{\varphi (n)})$ converges to some $\bar x \in Y$. Hence $\bar x$ is a cluster point of $(x_n)$.

  1. Every sequence in $X$ has a cluster point in $X$ $\implies$ Every open cover of $X$ has a finite subcover

Let $d$ be the metric on $X$.

First, we prove that $X$ is totally bounded. Suppose the contrary that $X$ is not totally bounded.

  • Then there exists $r>0$ such that $X \not \subseteq \bigcup_{k=0}^{m} \mathbb{B}\left(x_{k}, r\right)$ for any finite set $\{x_{0}, \ldots, x_{m}\} \subseteq K .$ In particular, there exists $x_{0} \in K$ such that $X \not \subseteq \mathbb{B} (x_{0}, r) .$ Thus there exists $x_{1} \in \left(\mathbb{B}(x_{0}, r)\right)^c$. Since $X \not \subseteq \left (\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)$, there exists $x_{2} \in \left(\mathbb{B}(x_{0}, r) \bigcup \mathbb{B}(x_{1}, r)\right)^c$. Continuing in this way and with Axiom of Countable Choice, there is a sequence $(x_n)$ in $X$ such that $x_{n+1} \in \left(\bigcup_{k=0}^n \mathbb{B}(x_{k}, r)\right)^c$. It follows that $x_{n+1} \notin \mathbb{B}(x_{k}, r)$ and thus $d(x_k, x_{n+1}) \ge r$ for all $k \le n$.

  • By our hypothesis, $(x_n)$ has a cluster point, i.e. there exists a subsequence $(x_{\psi(n)})$ of $(x_n)$ such that $x_{\psi(n)} \to \bar x$ as $n \to \infty$. It follows from $x_{\psi(n)} \to \bar x$ that there is $N \in \mathbb N$ such that $d(x_{\psi(n)},\bar x) < r/2$ for all $n \ge N$. Hence $d(x_{\psi(N)},\bar x) < r/2$ and $d(x_{\psi(N+1)},\bar x) < r/2$. It follows that $d(x_{\psi(N)}, x_{\psi(N+1)}) \le d(x_{\psi(N)},\bar x) + d(x_{\psi(N+1)},\bar x) < r/2 +r/2 = r$. This contradicts our construction of $(x_n)$. Hence $X$ is totally bounded.

Next, we prove that every open cover $\{O_i \mid i \in I \}$ of $X$ has a countable subcover.

  • Because $X$ is totally bounded, for each $n \ge 1$ there are finitely many $x^i_n$ such that $X = \bigcup_{i=0}^{k_n} \mathbb B (x_n^i , 1/n)$. Let $A = \bigcup_{n=1}^\infty \{x_n^0, \ldots, x_n^{k_n}\}$. Because $A$ is countable union of countable sets, $A$ is countable.

  • We define a mapping $f:A \times \mathbb Q_+ \to I$ by corresponding (with help from Axiom of Choice) $(a,r) \in A \times \mathbb Q_+$ with an $i \in I$ such that $\mathbb B(a,r) \subseteq O_i$ if such $i$ exists, otherwise $f(a,r) =i_0$ for some fixed $i_0 \in I$. Let $J=f[A \times \mathbb Q_+]$. Because $A,\mathbb Q_+$ are countable, $A \times \mathbb Q_+$ is countable and so is $J$.

  • For $x\in X$, there exists some $j \in I$ such that $x \in O_j$. Because $O_j$ is open, there is $r>0$ such that $\mathbb B(x,r) \subseteq O_j$. Then we choose some $a \in A$ such that $d(a,x) < r/2$ and some $r' \in \mathbb Q$ such that $d(a,x) <r' < r/2$. It follows that $x \in \mathbb B(a,r') \subseteq \mathbb B(x,r) \subseteq O_j$ by triangle inequality. By the construction of $f$, $x \in O_{f(a,r')}$. As such, $\{O_i \mid i \in J\}$ is a countable subcover of $X$.

Finally, we prove that $X$ is compact. Let $\{O_i \mid i \in I \}$ be an open cover of $X$.

  • We've just proved that $\{O_i \mid i \in I \}$ has a countable subcover $\{O_k \mid k \in \mathbb N\}$. Assume the contrary that $\{O_i \mid i \in I \}$ has no finite subcover. Then $X \not \subseteq \bigcup_{k =0}^n O_k$ for any $n \in \mathbb N$. As such, $X_n:=\bigcap_{k=0}^n O^c_{k} = \left (\bigcup_{k=0}^n O_{k} \right)^c \neq \emptyset$ for all $n \in \mathbb N$. Moreover, $X_{n+1} \subseteq X_n$ and $X_n$ is closed in $X$ for all $n$. On the other hand, $\{O_k \mid k \in \mathbb N\}$ is a subcover of $X$, so $\bigcup_{k=0}^\infty O_k =X$ or equivalently $\bigcap_{n=0}^\infty X_n = \bigcap_{k=0}^\infty O^c_{k} = \emptyset$.

  • By Axiom of Countable Choice, we define the sequence $(x_n)$ in $X$ recursively by $x_n \in X_n$ for all $n \in \mathbb N$. By hypothesis, $(x_n)$ has a cluster point $\bar x \in X$, i.e. there is a subsequence $(x_{\phi(n)})$ of $(x_n)$ such that $x_{\phi(n)} \to \bar x \in X$. It follows from our construction of $(x_n)$ that $x_{\phi(n)} \in X_N$ for all $n \ge N$. Moreover, $X_N$ is closed, so $\bar x \in X_N$. Because this is true for all $N$, we have $\bar x \in \bigcap_{n=0}^\infty X_n = \emptyset$. This is a contradiction. Hence $\{O_i \mid i \in I \}$ has a finite subcover.

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    $\begingroup$ That looks good to me. It goes into more detail than is needed, but this is something you learn with experience. One thing you'll discover as you advance in mathematics is that many arduous and difficult proofs become embarrassingly obvious when stated at the appropriate level of abstraction. Of course, Identifying that appropriate level of abstraction is itself a difficult thing to accomplish. $\endgroup$ – Paul Sinclair Oct 30 '19 at 23:19
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    $\begingroup$ I would like to thank you again for your kindness @PaulSinclair! Your comments really help me ^^ $\endgroup$ – Akira Oct 30 '19 at 23:26
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Assuming (iii) try to prove any open cover has a countable sub-cover first, i.e. some listable cover B_1,B_2,B_3,... . Then you can argue inductively that this can be reduced to a sub-cover B_1',B_2',B_3',... such that no B_i' is covered by the other sets. At this point you can pick x_1,x_2,x_3,... from each set but outside all the others. Either this sub-cover is finite or you have just constructed a sequence which can only accumulate to its limit x. However, x would have to be in one of the B_i'. This cannot be since only one of the x_j's is in this set.

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