A Kick in the Discovery

Let’s consider the metric space $([0,1], |\cdot|).$ Note, this is the subspace of the real line with the Euclidean metric. We claim that the following is an open cover of $[0,1].$

Indeed, each set is open in $[0,1].$ Furthermore, every $x\in [0,1]$ is contained in one of the intervals. Note that this open cover contains infinitely many open sets.

Next, we can see that this particular cover has a finite subcover,

So, as it turns out, we could cover $[0,1]$ with only four of the open sets.

Let’s consider again the metric space $([0,1], |\cdot|).$ We claim that the following is another open cover of $[0,1].$

Indeed, each set is open in $[0,1].$ Furthermore, every $x\in [0,1]$ is contained in one of the intervals. Note that this open cover contains infinitely many open sets.

Next, we can see that this particular cover has a finite subcover,

So, as it turns out, we could cover $[0,1]$ with only two of the open sets.

Let’s mix it up and consider the metric space $((0,1), |\cdot|).$

We claim that the following is an open cover of $(0,1).$

Indeed, each set is open in $(0,1).$ Furthermore, we can see that every $x\in (0,1)$ is contained in one of the intervals.

Note that this open cover contains infinitely many open sets. Furthermore, there is no finite subcover that covers all of $(0,1).$ Indeed, first note that

Thus, if there were a finite subcover, then we would be able to cover $(0,1)$ with only one interval $\left(\frac{1}{K},1\right).$ Where $K\in \N$ is the largest index that was contained in the finite subcover.

Thus, we see that some open covers have no finite subcovers.

Let’s continue with the metric space $((0,1), |\cdot|).$

We claim that the following is an open cover of $(0,1).$

Indeed, each set is open in $(0,1).$ Furthermore, we can see that every $x\in (0,1)$ is contained in one of the intervals.

Note that this open cover contains infinitely many open sets. Furthermore, there is a finite subcover this time. In particular,

covers the open unit interval.

Key Idea: We proceed by contradiction. So, suppose that we have an open cover of $[0,1]$ that does not have a finite subcover. Then `cut’ $[0,1]$ in half into $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right].$ At least one of those halves requires an infinite number of open sets in the cover to cover it. Focusing on that half-sized interval, we cut it in half and deduce that one of those quarter-sized intervals requires an infinite number of the open sets in the cover to cover it. We continue this process indefinitely. From there, we choose a point in each of those intervals and produce a Cauchy sequence that converges. From there, we deduce a contradiction.

Notation: Let $\ell([a,b]) = b-a$ be the length of the interval $[a,b].$ Similarly, for open intervals $\ell((a,b)) = b-a.$

Begining of Proof: To begin, let $\{U_i\}_{i\in \mathcal{I}}$ be an arbitrary open cover of $[0,1].$ Without loss of generality, we can suppose that every open set in the open cover is an open interval in $[0,1].$

We aim to show that there is a finite subcover. To this end, assume for the hope of a contradiction that there is no finite number of open sets in $\{U_i\}_{i\in \mathcal{I}}$ that cover $[0,1].$ Denote $I_0=[0,1],$ and cut it in half to get the intervals $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right].$ It follows that either $\left[0,\frac{1}{2}\right]$ or $\left[\frac{1}{2},1\right]$ requires an infinite number of open sets from $\{U_i\}_{i\in \mathcal{I}}$ to cover it. Indeed, if not, then both halves $\left[0,\frac{1}{2}\right]$ and $\left[\frac{1}{2},1\right]$ need only a finite number of open sets to cover, and hence the entire interval $[0,1]$ needs only a finite number of open sets to cover it.

Let’s assume, without loss of generality, that the interval $I_1=\left[\frac{1}{2},1\right]$ requires an infinite number of open sets from $\{U_i\}_{i\in \mathcal{I}}$ to cover it. Remark: Note that $I_1\subset I_0$ and $\ell(I_1)= \frac{1}{2}\ell(I_0)= \frac{1}{2}.$

Next, we cut $I_1=\left[\frac{1}{2},1\right]$ in half and run the same argument again on $\left[\frac{1}{2},\frac{3}{4}\right]$ and $\left[\frac{3}{4},1\right].$ Thus, we deduce that at least one of them requires an infinite number of open sets to cover it. Label this interval $I_2.$ Remark: Note that $I_2\subset I_1\subset I_0$ and $\ell(I_2)=\frac{1}{2}\ell(I_1)= \frac{1}{4}.$

We continue in the manner, cut $I_n$ in half, and choose the subinterval that requires an infinite number of open sets from our cover to cover it and label it $I_{n+1}.$ By induction, we see that $\ell(I_{n+1})= \frac{1}{2^{n+1}}.$

Thus, we have a sequence of nested closed intervals $I_{n+1}\subset I_n,$ each of which requires an infinite number of open sets to cover and is shrinking in length: $\ell(I_{n+1})= \frac{1}{2^{n+1}}.$

We now show that there is $\bigcap_{n}I_n=\{\alpha\}$ for some $\alpha\in [0,1].$ Indeed, for each of the intervals, $I_n = [x_n,y_n],$ choose the left endpoint of the interval in order to construct a sequence $(x_n)_{n\in \N}\subset [0,1].$ Furthermore, we can see that $(x_n)_{n}$ is Cauchy. Indeed, let $\varepsilon>0.$ There is some $N\in \N$ such that $\frac{1}{2^N}<\varepsilon.$ Consequently, for all $m,n\in \N$ such that $m,n>N$ we have $x_m,x_n\in I_N$ (since $I_{n}\subset I_N$ and $I_{m}\subset I_N$ ) and thus $|x_n-x_m|<\frac{1}{2^N}<\varepsilon.$ Therefore, we know that $(x_n)_{n}$ converges to some $\alpha\in [0,1].$ In particular, $\alpha\in I_n$ for all $n\in \N.$ Therefore, $\alpha\in \bigcap_{n}I_n.$

To see that $\bigcap_{n}I_n=\{\alpha\},$ suppose that there was some $\beta\neq \alpha$ such that $\beta \in \bigcap_{n}I_n.$ Then, let $\varepsilon = |\alpha-\beta|>0.$ However, there is some $N\in \N$ such that $\frac{1}{2^N}<\varepsilon,$ and therefore both $\alpha$ and $\beta$ we cannot be in $I_N.$ Thus, a contradiction and therefore $\bigcap_{n}I_n=\{\alpha\}.$

Since, $\alpha\in [0,1]$ it follows that there is some open set $U_{i_{\alpha}}$ in the cover such that $\alpha \in U_{i_{\alpha}}.$

Let $\ell(U_{i_{\alpha}}) = L>0.$ There is some $N\in \N$ such that $\frac{1}{2^N}<L$ and consequently, $I_N\subset U_{i_{\alpha}}.$ However, the intervals $I_n$ were constructed so that each of them required an infinite number of open sets to cover them. However, $I_N$ only requires one open set, namely $U_{i_{\alpha}}$ to cover it. This is our contradiction.

We conclude that $\{U_i\}_{i\in \mathcal{I}}$ has a finite subcover of $[0,1].$ Hence $([0,1], |\cdot|)$ is compact.

Let $(X,d)$ be a finite metric space and let $\{U_i\}_{i\in \mathcal{I}}$ be an open cover of $X.$ It follows that each $x \in U_{i_x}.$ Thus, $\{U_{i_x}\;:\;x\in X\}$ forms a finite subcover, since $X$ is a finite set.

Let $(x_n)_{n\in \N}$ be a sequence of real numbers that is bounded below by $a$ and bounded above by $b.$ I.e. $(x_n)_{n\in \N}\subset [a,b].$ We claim that $(x_n)_{n\in \N}$ contains a monotone subsequence. Note, this proves the theorem since, by the monotone convergence theorem, a bounded monotone sequence converges.

Terminology: Before we begin, we call an element $x_{k}$ from our sequence a tiptop element or point if $x_{m}\leq x_k$ for all $m>k.$ That is, every point further into the sequence is less than or equal to $x_{k}.$

There are two cases we need to consider (i) there are an infinite number of tiptop points, or (ii) there are only a finite number of tiptop points.

In case (i), the sequence of tiptop points forms a monotone decreasing subsequence. Thus, we are done.

In case (ii), let $x_{N}$ be the last tiptop point. It follows that for all $x_{k}$ with $k\geq N,$ there is some $x_{m}$ such that $x_{k} \leq x_m.$ I.e., there is a monotone increasing subsequence! Thus, we are done again!

Let $(X,d)$ be a metric space.

Forward: $X$ is compact $\implies$ $X$ is sequentially compact.

Let $X$ be compact and let $(x_n)_{n\in \N}\subset X$ be a sequence of elements from $X.$ We aim to show that there is a subsequence $(x_{n_k})_k$ that converges to some $x_0\in X.$ To this end, consider the following claim:

Claim 1: There is some point $x_0\in X$ such that for all $\varepsilon>0,$ the open ball $B(x_0\,,\,\varepsilon)$ contains an infinite number of elements from the sequence $(x_n)_n.$ That is, there is an infinite number of $n\in \N$ such that $x_n\in B(x_0\,,\,\varepsilon).$

We continue by contradiction. Suppose that for all $x\in X$ there is some $\varepsilon(x)>0,$ such that each open ball $B(x\,,\,\varepsilon(x))$ contains only a finite number of elements from $(x_n)_n.$ Next, note that the collection $\{B(x\,,\,\varepsilon(x))\;:\; x\in X\}$ forms an open cover of $X.$ It follows by compactness, there is a finite collection of open balls, those open balls $\{ B(p_1\,,\,\varepsilon(p_1))\;,\; B(p_2\,,\,\varepsilon(p_2))\,,\, \cdots , B(p_N\,,\,\varepsilon(p_N))\}$ that cover $X.$

Now, we have reached our contradiction. Indeed, since each open ball in the finite cover contains only a finite number of elements from $(x_n)_{n}$ we conclude that there are only a finite number of elements (or natural numbers) in $(x_n)_{n}.$ Hence, claim 1 is true.

Now, using claim 1, we construct a convergent subsequence of $(x_n)_{n}.$

Let $\varepsilon_n = 2^{-n}>0.$ Then, let $x_{n_1} \in B(x_0,\varepsilon_1).$ Which exists by claim 1. Next, since there is an infinite number of elements from $(x_n)_{n}$ in $B(x_0,\varepsilon_2),$ there is some $x_{n_2} \in B(x_0,\varepsilon_2)$ such that $n_2>n_1.$ We continue inductively. Suppose that we have determined $x_{n_2}, \cdots , x_{n_k},$ then we let $x_{n_{k+1}}\in B(x_0,\varepsilon_{k+1})$ such that $n_{k+1}>n_k.$

We claim that the subsequence $(x_{n_k})_k$ converges to $x_0.$ Indeed, let $\varepsilon>0.$ It follows that there is some $N\in \N$ such that $\varepsilon_N = 2^{-N}<\varepsilon.$ Consequently, for all $n>N$ we have $x_n\in B(x_0,\varepsilon_N),$and thus $d(x,x_0)<\varepsilon.$

We prove this by contradiction. That is, suppose that for all $\varepsilon>0,$ there is some open ball $B$ with radius $\varepsilon$ that is not contained in any single open set $U_i.$ In particular, for all $\frac{1}{n}>0,$ there is some open ball $B_n$ with radius $\frac{1}{n}$ that is not contained in any open set $U_i.$

We focus on the centers of the balls $B_n$ and deduce a contradiction, so let $p_n\in B_n$ be the ball’s center.

Consider the sequence of centers $(p_n)_n\subset X.$ By sequential compactness, there is some convergent subsequence denoted $(p_{n_k})_k.$ Let $p_{n_k}\rightarrow p\in X$ as $k\rightarrow \infty.$ Since $\{U_i\}_{i\in \mathcal{I}}$ covers $X,$ there is some $U_{i_p}$ that contains $p.$ Since $U_{i_p}$ is open, there is some $\delta>0$ such that $B(p,\delta)\subset U_{i_p}.$

Next, since $p_{n_k}\rightarrow p$ as $k\rightarrow \infty$ there is some $K\in \N$ such that for all $k>K$ we have $d(p_{n_k},p)<\delta/2.$ Furthermore, there is some $M\in \N$ such that for all $m>M$ we have $\frac{1}{n_m}<\delta/2.$ Let $\mathcal{K} = \max{\{K,M\}}.$ We claim that for all $k>\mathcal{K}$ we have $B_{n_k}\subset B(p,\delta).$ Indeed, let $x\in B_{n_k} =B\left(p_{n_k},\frac{1}{n_m}\right).$ It follows

Thus, we see that $B_{n_k}\subset B(p,\delta)\subset U_{i_p}.$ This contradicts the assumption that $B_{n_k}$ was not contained in any open set in the cover. Thus, proving Lemma 1.

Key Idea: We will show that if $X$ is not totally bounded, then there would be a sequence that has no convergent subsequence. Thus contradicting that $X$ is sequentially compact.

We proceed by contradiction. Suppose that there were some $\varepsilon>0$ so that there is no finite subcover of $\{B(p\,,\,\varepsilon)\;:\; p\in X\}.$ Then, we claim:

Claim: There is a sequence of points $p_1,p_2,\cdots \in X$ such that $d(p_i,p_j)\geq \varepsilon$ for all $i\neq j.$

To see this, we construct $p_1,p_2,\cdots \in X$ inductively.

Choose any $p_1\in X.$ Suppose that we have chosen $p_2,\cdots , p_n\in X.$ It follows that there is some $p_{n+1}$ such that $d(p_{n+1},p_i)\geq \varepsilon$ for all $1\leq i\leq n.$ If not, then for all $p_{n+1}\in X$ there is some $1\leq i\leq n$ such that $d(p_{n+1},p_i)< \varepsilon.$ However, this implies $p_{n+1}\in B(p_i,\varepsilon),$ and hence $X\subset \bigcup_{i=1}^n B(p_i,\varepsilon).$ This contradicts the assumption that there was no finite subcover of $\{B(p\,,\,\varepsilon)\;:\; p\in X\}.$ Hence, we can find some $p_{n+1}$ such that $d(p_{n+1},p_i)\geq \varepsilon$ for all $1\leq i\leq n.$ Thus our claim holds.

By our claim, the sequence $(p_n)_n\in X$ is such that $d(p_i,p_j)\geq \varepsilon$ for all $i\neq j.$ Thus, $(p_n)_n$ has no convergent subsequence since no subsequence can be Cauchy. Indeed, since $d(p_i,p_j)\geq \varepsilon$ for all $i\neq j,$ the sequence never gets arbitrarily close to itself. This contradicts the fact that $X$ is sequentially compact. Thus, we must have a finite subcover of $\varepsilon$-balls: $\{B(p_1\,,\,\varepsilon)\;,\; \cdots, B(p_n\,,\,\varepsilon)\},$ that covers $X.$

Backward: $X$ is sequentially compact$\implies$$X$ is compact.

Let $X$ be sequentially compact and let $\{U_i\}_{i\in \mathcal{I}}$ be an open cover of $X.$ We aim to show that there is a finite subcover. Next, let $\varepsilon>0$ be such that Lemma 1 holds. That is, every $\varepsilon$-ball $\{B(p\,,\,\varepsilon)\;:\; p\in X\}$ is contained in some open set $U_{i_p}\in \{U_i\}_{i\in \mathcal{I}}$ in the open cover. I.e., for all $p\in X,$ there is some $U_{i_p}\in \{U_i\}_{i\in \mathcal{I}}$ such that $B(p\,,\,\varepsilon)\subset U_{i_p}.$

By Lemma 2, there is a finite subcover of the cover of $\varepsilon$-balls, $\{B(p_1\,,\,\varepsilon)\;,\; \cdots, B(p_n\,,\,\varepsilon)\}$ that covers $X.$ However, this implies that there is finite subcover $\{U_{i_1},\cdots, U_{i_n}\}$ covers $X.$ Indeed, since every $B(p_k\,,\,\varepsilon)\subset U_{i_{p_k}}.$Thus concluding the proof.

$\bigstar$ The rest of (2) $\implies$(3)

Let $(X,d)$ be a sequentially compact metric space. We aim to prove that $X$ is totally bounded and complete. By Lemma 2 $X$ is totally bounded. To prove completeness, consider some Cauchy sequence $(x_n)_{n\in \N}\subset X,$ we need to show that $x_n \rightarrow x$ for some $x\in X.$

To begin, by sequential compactness, there is some convergent subsequence $(x_{n_k})_{k}$ with limit $x\in X.$ We claim that $x_n \rightarrow x$ too.

Let $(x_n)_n$ is Cauchy, there is some $N\in \N$ such that for all $n,m>N$ we have $d(x_n,x_m)<\varepsilon/2.$ Furthermore, there is some $K\in \N$ such that for all $k>K$ we have $d(x_{n_k},x)<\varepsilon/2.$ Consequently, for all $n>\max{\{n_K,N\}}$ we have

$\bigstar$ Now we aim to show (3) $\implies$(2)

To this end, let $X$ be totally bounded and complete, and let $(x_n)_{n\in \N}\subset X.$ We will show that there is a convergent subsequence of $(x_n)_{n}.$ Here’s the key idea. Let $\varepsilon_n = 1/n>0.$ We cover $X$ with finitely many $\varepsilon_1$-balls. We can do this since $X$ is totally bounded. It follows that there is an open ball that contains an infinite number of elements from $(x_n)_{n}.$ Now ignore all other elements from the original sequence. Cover $X$ with finitely many $\varepsilon_2$-balls. Note, from the elements remaining from the original sequence, there is one open $\varepsilon_2$-ball that contains an infinite number of those! We continue this process and produce a subsequence that is Cauchy. By completeness, it converges. We will make this precise, and in turn change some of the details for convenience.

This type of proof is called a diagonal proof and is always accompanied with a picture that looks like the following.

We will construct the subsequence as follows. First, let $(x_n^{(1)})_{n}$ be our original sequence $(x_n)_{n}.$ For $m\geq 2,$we let $(x_n^{(m)})_{n}$ have the following properties:

• $(x_n^{(m)})_{n}$ is a subsequence of $(x_n^{(m-1)})_{n}$ and
• $(x_n^{(m)})_{n}$ is contained in a ball with radius $\varepsilon_m=1/m.$

We use induction. We alrady have $(x_n^{(1)})_{n} =(x_n)_n.$ Let $m\geq 2$ and suppose that we have defined $(x_n^{(m)})_{n}$ for $k<m.$ Cover $X$ with finitely many $\varepsilon_m$-balls. It follows that there are an infinite number of elements from $(x_n^{(m-1)})_{n}$ in at least one of the $\varepsilon_m$-balls. Denote this particular ball $B_m.$ We then let $x_1^{(m)}$ be the first element from $(x_n^{(m-1)})_{n}$ that is in $B_m.$ Then we let the $x_2^{(m)}$ be the second element from $(x_n^{(m-1)})_{n}$that is in $B_m.$ Etc.

We now construct a diagonal sequence from the $(x_n^{(m)})_{n}.$ This is where the picture comes in. Let $(a_n)_{n}$ be the sequence such that $a_n=x_n^{(n)}.$ We can see that $(a_n)_{n}$ is indeed a subsequence of $(x_n)_{n}.$ Furthermore, we see that $(a_n)_{n}$ is Cauchy. Indeed, let $\varepsilon>0.$ There is some $N\in \N$ such that $\varepsilon_n = 1/n<\varepsilon/2.$ Thus, for all $m,n>N,$ we have $(a_n)_{n>N}$ being a subsequence of $(x_n^{(N)})_{n}.$ Thus, $d(a_n,a_m)<2\varepsilon_n<\varepsilon.$ This concludes the proof.

Let $(X,d_X)$ be a compact metric space and let $f:X\rightarrow Y$ be continuous.

Let $\varepsilon>0.$ We aim to find some $\delta>0$ such that for all $a,b\in X$ where $d_X(a,b)<\delta$ we have $d_Y(f(a),f(b)<\varepsilon.$

To begin, since $f$ is continuous on $X,$ for all $p\in X$ there is some $\delta_p>0$ so that $d_X(p,q)<\delta_p$ implies $d_Y(f(p),f(q))<\varepsilon/2.$ Note that $\delta_p$ depends on the point $p\in X$ chosen. Next, we cover $X$ with $\delta_p/2$-balls. I.e., we’re using the open cover, $\{B_X(p,\delta_p/2) \}_{p\in X}$ where $B_X(p,\delta_p/2) = \{q\in X\;:\;d_X(p,q)<\delta_p/2\}$ is an open ball in $(X,d_X).$ Since $(X,d_X)$ is compact, there is a finite subcover denoted $\{B_X(p_1,\delta_{p_1}/2),\cdots, B_X(p_n,\delta_{p_n}/2) \}.$

Let $\delta:=\frac{1}{2}\min{\{\delta_{p_i}\;:\;1\leq i\leq n\}}>0.$ We claim this does the job. Indeed, for all $a,b\in X$ where $d_X(a,b)<\delta$ we have $a,b\in B(p_i,\delta_{p_i}/2)$ for some $1\leq i\leq n.$ Indeed, note that $a\in B(p_i,\delta_{p_i}/2)$ for some $1\leq i\leq n$ since the open balls cover $X.$ Furthermore, we have $d_X(b,p_i)\leq d_X(a,b)+d_X(a,p_i) <\delta + \frac{\delta_{p_i}}{2}\leq\delta_{p_i}.$ Hence $a,b\in B(p_i,\delta_{p_i}/2)$ as claimed.

Finally, by definition of $\delta,$ $d_X(a,b)<\delta$ implies

Thus concluding the proof.

Let $(X,d_X)$ be a compact metric space and let $f:X\rightarrow Y$ be continuous. Since we aim to show that $f(X)$ is compact, we let $\{V_i\}_{i\in \mathcal{I}}$ be an open cover of $f(X).$ Note, $V_i\subset Y.$ We need to find a finite subcover.

First, since $f$ is continuous, the preimages of the open sets in the cover are open. Thus, $\{f^{-1}(V_i)\}_{i\in \mathcal{I}}$ is an open collection of open sets in $X.$ We claim that $\{f^{-1}(V_i)\}_{i\in \mathcal{I}}$ is an open cover of $X.$ Indeed, let $x\in X$ and observe that $f(x)\in f(X).$ Thus, $f(x)\in V_{i_0}$ for some $i_0$ hence, $x\in f^{-1}(V_{i_0}).$ Consequently, $X\subset \bigcup_{i\in \mathcal{I}}f^{-1}(V_i).$

Next, by the compactness of $X,$ there is a finite subcover $\{f^{-1}(V_{i_k})\}_{k=1}^n$ that covers $X.$ We claim that the corresponding $\{V_{i_k}\}_{k=1}^n$ cover $f(X).$ Indeed, let $f(x)\in f(X).$ It follows that $x\in f^{-1}(V_{i_k})$ and thus $f(x)\in V_{i_k}.$

Thus concluding the proof.