3.6Limit Point Compactness

Proof.

Suppose $X$ is compact. We seek to show if $A \subseteq X$ is infinite, then it contains a limit point in $X$. We prove by contrapositive: that if $A$ has no limit points, then it is finite. Indeed, take $A$ to have no limit points. Then $A$ is trivially closed, since it contains all its limit points. Since $A$ is a closed subset of a compact space, we have that $A$ is compact as well. For each $a \in A$, note that $U_a = \{a\}$ forms an open neighborhood of $a$ in $A$, and so $\{U_a\}_{a \in A}$ is an open cover for $A$. Since $A$ is compact, we have that $\{U_k\}_{k = 1}^n$ forms a finite subcover, meaning $A$ has finitely many elements. ❦

Proof.

$(1) \implies (2)$: We have proven this already.

$(2) \implies (3)$: Suppose $X$ is limit point compact. Let $x_n$ be an arbitrary sequence in $X$, and let $A = \{x_n\}_{n \in \bZ_+}$. If $A$ is finite, then there is one number that repeats a countably infinite number of times in the sequence, so choose this as a subsequence and note it converges. Otherwise, if $A$ is infinite, then by hypothesis we have a limit point $x$. For each $k \in \bZ_+$, let $n_k$ both be greater than all $n_j$ for $j < k$ and be such that $x_{n_k} \in B(x, 1/k)$. Then $(x_{n_k}) \to x$, and so $X$ is sequentially compact.

$(3) \implies (1)$: Suppose $X$ is sequentially compact. We prove that the Lebesgue covering lemma applies to $X$, then we prove that finitely many $\epsilon$-balls can cover $X$, then we finally prove $X$ is compact.

Let $\mathcal{A}$ be an arbitrary open cover of $X$. For the sake of contradiction, suppose the Lebesgue covering lemma doesn't apply to $X$. Then, for each $n \in \bZ_+$, there exists a nonempty set $C_n$ whose diameter is less than $1/n$ and does not lie in a single member of $\mathcal{A}$. Choosing $x_n \in C_n$ for each $n$ and forming a sequence, we use our hypothesis to get a convergent subsequence $(x_{n_k}) \to a$. Note that $a$ belongs to some member $A$ of the open cover $\mathcal{A}$, and since $A$ is open, there is some $\epsilon > 0$ such that $B(a, \epsilon) \subseteq A$. We may choose an index $i$ large enough such that $1/n_i < \epsilon/2$, and since $\diam(C_{n_i}) < \epsilon/2$, we have that $C_{n_i} \subseteq B(x, \epsilon/2)$; moreover, we may choose $i$ large enough such that $d(x_{n_i}, a) < \epsilon/2$. For any point $c \in C_{n_i}$, note that

and so $C_{n_i} \subseteq B(a, \epsilon)$. Thus $C_{n_i} \subseteq A$, contrary to assumption.

Suppose, for the sake of contradiction, that $X$ cannot be covered by finitely many $\epsilon$-balls, for a fixed $\epsilon > 0$. Let $x_1 \in X$, then note $B(x_1, \epsilon) \ne X$, otherwise $X$ could be covered by finitely (namely one) $\epsilon$-ball. Thus choose $x_2 \in X \setminus B(x_1, \epsilon)$, and note $B(x_1, \epsilon) \cup B(x_2, \epsilon) \ne X$ for the same reason as before. In general, given distinct points $x_1, x_2, \dots, x_n$ of $X$, choose $x_{n + 1}$ such that $x_{n + 1} \notin \bigcup_{k = 1}^n B(x_k, \epsilon)$. By construction, we have that $d(x_{n + 1}, x_i) \ge \epsilon$ for every $i \in \{1, \dots, n\}$, so there is no chance of $(x_n)$ having a convergent subsequence, contrary to the fact that $X$ is sequentially compact.

Finally, we show $X$ is compact. Let $\mathcal{A}$ be an arbitrary open cover ofr $X$. Since $X$ is sequentially compact, it has a Lebesgue number $\delta$. Choose $\epsilon = \delta/3$, then note that we may cover $X$ with finitely many $\epsilon$-balls. Each of these balls has a diameter of $2\delta/3$, so each ball fits in one member of the open cover. There are finitely members of the cover that contain all the balls, which cover $X$, and so $X$ is compact. ❦

Proof.

We seek to find some infinite subset $A \subseteq [0, 1]^\omega$ with no limit points. Consider the set of all standard basis vectors of $\bR^\omega$—that is, the set $A = \{e_n\}_{n \in \bZ_+}$, where $e_n: \bZ_+ \to [0, 1]$ is defined by $e_n(j) = \delta_{n, j}$. Then, for any $e_j, e_k$ for $j \ne k \in \bZ_+$, we have that

For the sake of contradiction, suppose $x \in [0, 1]^\omega$ is a limit point. Then $B(x, 1/3) \cap [A \setminus \{x\}] \ne \emptyset$. In fact, $B(x, 1/3)$ intersects $A$ infinitely many times outside of $x$. If $B(x, 1/3) \cap [A \setminus \{x\}] = \{e_1, \dots, e_n\}$, then we could take $r = \min\{d(x, e_k)\}_{k = 1}^n$ and choose $r' < r$ so that $B(x, r') \cap [A \setminus \{x\}] = \emptyset$ (and so $x$ not a limit point), a contradiction. Thus let $e_1, e_2 \in B(x, 1/3) \cap [A \setminus \{x\}]$. Then note

which means $e_1 \notin B(x, 1/3)$, a contradiction. ❦

Solution.

1. No. Let $X = \bZ_+ \times Y$, where $Y = \{a, b\}$ with the indiscrete topology. Then define $f: X \to \bZ_+$ (where $\bZ_+$ takes its subspace topology from $\bR$) by $(n, *) \mapsto n$. Note that any singleton subset $\{n\} \subseteq \bZ_+$ is a basic, open set, so showing $f^{-1}(\{n\})$ is open guarantees continuity. Indeed, the inverse image is $\{n\} \times Y$, which is an open subset of $X$. Of course, $\bZ_+$ is not limit point compact in its metric topology, completing the disproof.

2. Yes. Suppose $U \subseteq A$ is infinite. Since $U \subseteq X$, we have that $U$ has a limit point $x \in X$. Since $A$ contains all its limit points, it follows that $x \in A$, and so $A$ is limit point compact.

3. No. For our counterexample, we want $A$ to have a limit point in $Z \setminus X$, meaning $X$ is not closed (it does not contain all of its limit points). Consider $X = S_{\Omega}$, which is limit point compact, and $Z = \bar{S}_{\Omega}$, which is Hausdorff (all order topologies are Hausdorff). Note the subset $X \subseteq X$ has a limit point $\Omega$, but $\Omega \notin X$, and so $X$ is not closed.

Proof.

$(\Longrightarrow)$: Suppose $X$ is $T_1$ and countably compact: we proceed by contrapositive. Suppose $A \subseteq X$ has no limit points: we show that $A$ is finite. Indeed, note that $A$ contains all of its limit points, and so it is closed and contains only isolated points. Thus singleton subsets $\{*\} \subseteq A$ are open in $A$. For the sake of contradiction, suppose $A$ is infinite. Then we may choose a countably infinite set of distinct points $[A' = \{x_n\}_{n \in \bZ_+}] \subseteq A$. Consider the countably infinite cover of $X$ formed by taking each singleton subset of $A'$ and adjoining it to $X \setminus A'$. By countable compactness of $X$, we take a finite subcover; discarding the $X \setminus A'$ element, we get a covering of $A$ by finitely many singletons, a contradiction. Thus $A$ is finite, meaning $X$ is limit point compact.

$(\Longleftarrow)$: Suppose $X$ is $T_1$ and limit point compact: we show $X$ is countably compact by contradiction. Suppose $X$ is not countably compact, and take a countable open cover $\mathcal{U} = \{U_n\}_{n \in \bZ_+}$ of $X$. Note that there is no finite subcover, so for each $n \in \bZ_+$, choose $x_n \notin U_1 \cup \cdots \cup U_n$. Since $X = \bigcup U_n$, the set $A = \{x_n\}_{n \in \bZ_+}$ is infinite. By limit point compactness, $A$ has a limit point $x$. Since $\mathcal{U}$ covers $X$, $x \in U_k$ for some $k \in \bZ_+$. Since $X$ is $T_1$, the neighborhood $U_k$ of the limit point $x$ must contain infinitely many points of $A$. However, by construction, $x_n \notin U_k$ for all $n \ge k$, implying $U_k \cap A \subseteq \{x_1, \dots, x_{k-1}\}$. This finite intersection contradicts $x$ being a limit point. Thus, $X$ is countably compact. ❦

Solution.

1. Suppose $f(x) = f(y)$ for some $x, y \in X$. Then note $0 = d(f(x), f(y)) = d(x, y)$, meaning $x = y$ and so $f$ is injective. Next, choose $\epsilon > 0$, and consider the ball $O = B(x, \epsilon) \subseteq X$, a basic open set. For any $y \in O$, note that $d(f(x), f(y)) = d(x, y)$, so $f(y) \in B(f(x), \epsilon)$. Indeed, $f(O) = B(f(x), \epsilon) \cap f(X)$, meaning $f^{-1}|_{f(X)}$ is continuous. Showing $f: X \to f(X)$ is continuous follows almost exactly the same idea. Thus $f$ is a homeomorphism onto its image, i.e. is an embedding.

2. By continuity of $f$, we have that $f(X)$ is compact. Note that $f(X) \subseteq X$ is a compact subspace of a Hausdorff space, and so $f(X)$ is closed. For the sake of contradiction, suppose $a \in X \setminus f(X)$. Since $a$ belongs to an open set, there is some $\epsilon > 0$ such that $B(a, \epsilon) \cap f(X) = \emptyset$. Note then that $d(a, f(a)) \ge \epsilon$. Moreover, $d(f(a), f(f(a))) = d(a, f(a)) \ge \epsilon$, where the equality is by hypothesis. Indeed, for $x_1 = a$ and $x_{n + 1} = f(x_n)$, we have that $d(x_m, x_n) \ge \epsilon$ for every $m \ne n$. Note the sequence $\{x_n\}_{n \in \bZ_+ \setminus \{1\}} \subseteq f(X)$ has no convergent subsequence, and so $f(X)$ is not sequentially compact, a contradiction (since compactness and sequential compactness are equivalent in metrizable spaces). Thus $X \setminus f(X) = \emptyset$, and so $f(X) = X$, meaning $f$ is a surjective embedding and thus a homeomorphism.

Solution.

1. Let $A_n = f^n(X)$, where $f^n$ denotes applying the function $n$ times. Note that $f: X \to X$, and so $f(X) \subseteq X$ since $f(X)$ is a subset of the codomain. We will show $f(f(X)) \subseteq f(X)$, and then by induction, it will follow that $A_{n + 1} \subseteq A_n$ for any $n \in \bZ_+$. If $y \in f(f(X))$, there is some $x \in f^{-1}(\{y\}) \in f(X) \subseteq X$, and so $y = f(x) \in f(X)$ as desired.

   Note that contractions are Lipschitz and thus continuous. Since $X$ is compact, each $A_n$ is compact, and as a subspace of a metric (thus Hausdorff) space, each $A_n$ is closed. Since $A_n$ forms a sequence of nested closed sets, by compactness of $X$ we have that there is some $x^* \in \bigcap_{n \in \bZ_+} A_n$. We show $x^*$ is our desired fixed point.

   Let $A = \bigcap_{n \in \bZ_+} A_n$, and note $x^* \in A$. Note that all potential fixed points must live in $A$, so to show $x^*$ is unique, we show $A$ has only one point. Since $A_{n + 1} \subseteq A_n$, we have that $\diam(A_{n + 1}) \le \diam(A_n)$. Since $X$ is a compact metric space, we have that $X$ is bounded, meaning $\diam(X) = M$ for some $M \in \bR$. Then $\diam(A_n) = \diam(f^n(X)) \le C^n M$, and as $n \to \infty$, we have $\diam(A_n) \to 0$, implying $\diam(A) = 0$. Thus $A = \{x^*\}$.

   Finally, to show $x^*$ is a fixed point, we show $f(A) \subseteq A$. If $y \in f(A)$, then $y = f(x^*)$. Since $x^* \in A_n$ for each $n \in \bZ_+$, we have that $f(x^*) \in f(A_n) = A_{n + 1} \subseteq A_n$ for each $n \in \bZ_+$. Thus $y \in A$, meaning $f(\{x^*\}) \subseteq \{x^*\}$. This forces $f(x^*) = x^*$, meaning $x^*$ is a fixed point of $f$, and by the previous argument, this fixed point is unique.

2. Let $A_n = f^n(X)$ and $A = \bigcap_{n \in \bZ_+} A_n$. As in the previous part, $f$ is continuous (since it is a shrinking map, it is 1-Lipschitz), each $A_n$ is closed and nested, and $A$ is a nonempty compact set with $f(A) \subseteq A$. We show $A \subseteq f(A)$.

   Let $x \in A$. For each $n \in \bZ_+$, since $x \in A_{n+1} = f^{n+1}(X)$, there exists $x_n \in X$ such that $x = f^{n+1}(x_n)$. Define $y_n = f^n(x_n)$, noting $y_n \in A_n$ and $f(y_n) = x$ for all $n$. Since $X$ is sequentially compact, $(y_n)$ has a convergent subsequence $(y_{n_k}) \to a \in X$. For any fixed $m \in \bZ_+$, the tail of the subsequence lies entirely in the closed set $A_m$, so $a \in A_m$. Thus $a \in A$. By continuity of $f$, $f(a) = \lim_{k \to \infty} f(y_{n_k}) = \lim_{k \to \infty} x = x$. Thus $x \in f(A)$, so $A = f(A)$.

   We now show $\diam(A) = 0$. Since $A$ is compact, if $\diam(A) > 0$, the continuous function $d$ attains a maximum on $A \times A$, so there exist $p, q \in A$ such that $d(p, q) = \diam(A)$. Since $A = f(A)$, we can choose $u, v \in A$ such that $f(u) = p$ and $f(v) = q$. Since $p \ne q$, we must have $u \ne v$. Because $f$ is a shrinking map,

   $$\diam(A) = d(p, q) = d(f(u), f(v)) < d(u, v) \le \diam(A),$$

   which is a contradiction. Thus $\diam(A) = 0$, meaning $A = \{x^*\}$.

   Since $f(A) = A$, we have $f(x^*) = x^*$. If $y$ is another fixed point, $d(x^*, y) = d(f(x^*), f(y)) < d(x^*, y)$, a contradiction. Thus $x^*$ is unique.