MA3209 Past Final Papers Attempts

MA3209 Metric and Topological Spaces is a mathematics course offered in the National University of Singapore (NUS) that revolves around general topology, with a focus on metric spaces.

This page presents my own attempt on (a) past-year final exam paper(s) of this course.

Academic Year 2023/2024 Semester 2

1. Indicate if the following statements are true or false.

   (a) If $f : X \to Y$ is continuous and $X$ is connected, then $f(X)$ is connected.

   (b) The metric completion of a topological space is compact.

   (c) If $X$ is a normal topological space and $C \subseteq X$ is a closed subspace, then $C$ is normal.

   (d) Every metric space is second countable.

   (e) Let $X$ be a set and equip $\mathbb{R}$ with the standard metric. The set $\mathcal{B}(X,\mathbb{R})$ of bounded functions on $X$, when equipped with the supremum metric, is complete.

(a) True. See Theorem 23.5 in Chapter 3 of Munkres.

(b) False. The set of real numbers $\mathbb{R}$ is a metric completion of the set of rational numbers $\mathbb{Q}$ under the standard metric, but $\mathbb{R}$ is not compact.

(c) True. See Exercise 32.1 in Chapter 4.

(d) False. the collection of sequences of real numbers $\mathbb{R}^\omega$ in the uniform topology is not second-countable.

(e) True. See Theorem 43.6 in Chapter 7. Note that $\mathbb{R}$ with the standard metric is complete and the uniform metric is equivalent to the supremum metric in $\mathcal{B}(X,\mathbb{R})$.

2. Equip $\{0,1\}$ with the discrete topology, and equip $\{0,1\}^{\mathbb{R}}$ with the induced product topology. Let $X \subseteq \{0,1\}^{\mathbb{R}}$ be the subset defined by

   $$X = \{f \in \{0,1\}^{\mathbb{R}} : f^{-1}(\{1\}) \text{ is at most countable}\}.$$

   (a) Show that $X$ is sequentially compact.

   (b) Show that $X$ is not compact.

(a) Note that $\{0,1\}$ is finite and thus compact. It is also metrizable since it is equipped with the discrete topology.

Let $A = \bigcup_{n \in \mathbb{Z}_+}f_n^{-1}(\{1\}) \subseteq \mathbb{R}$, then $A$ is countable since it is a countable union of countable sets. Since arbitrary products of compact spaces are compact and countable products of metrizable spaces are metrizable, it follows (by Theorem 28.2) that $\{0,1\}^A$ is sequentially compact.

Let $(f_n)$ be a sequence in $X$. Define a projection map $m : X \to \{0,1\}^A$ by $x \to x|_{\{0,1\}^A}$, i.e. obtaining an element in $\{0,1\}^A$ by only taking coordinates of $x$ that are indexed by elements in $A$. It follows that there is a subsequence $m(f_{n_k})$ that converges to, say, $m(f)$.

By filling in the coordinates indexed by elements in $\mathbb{R} \backslash A$ with zeroes, we obtain a subsequence $(f_{n_k})$ of $(f_n)$ that converges to $f$, and hence $X$ is sequentially compact.

(b) For each $r \in \mathbb{R}$, define $B_r = \{f \in X : f(r)=0\}$. It follows that $\{B_r\}_{r \in \mathbb{R}}$ is an open cover with no finite subcover, and thus $X$ is not compact.

To see this, note that each $B_r$ is open because it is generated by $\prod_{\alpha \in \mathbb{R}}U_\alpha$, where $U_\alpha \subseteq \{0,1\}$ is open in $\{0,1\}$, $U_\alpha=\{0,1\}$ except for finitely many values of $\alpha$, and every subset of $\{0,1\}$ is open. $\{B_r\}_{r \in \mathbb{R}}$ also covers $X$ because each $B_r$ is a subset of $X$ and $r$ ranges over all real number indices. $\{B_r\}_{r \in \mathbb{R}}$ also has no finite subcover because for every element in $X$, there are at least uncountable zeroes as its entries.

3. Let $X$ be a topological space.

   (a) Show that if $X$ is locally path connected, then the quotient topology on the set of connected components of $X$ is discrete.

   (b) Find an example of a space $X$ where the quotient topology on the set of connected components of $X$ is not discrete.

(a) Since $X$ is locally path connected, the connected components and path components of $X$ are the same¹. For every open set $U$ of $X$, in particular $X$ itself, each path (connected) component of $U$ is open in $X$². By definition, the quotient topology on the set of connected components of $X$ (which we denote here by $A$) is induced by a surjective map $p : X \to A$ that is a quotient map, and is defined by the subsets $V$ of $A$ such that $p^{-1}(V)$ is open in $X$. It follows that every subset $V$ of $A$ can be expressed as $\bigcup_{v \in V}\{v\}$, where each $v$ is a connected component of $X$, which implies that $p^{-1}(\{v\})$ is open in $X$ and thus $p^{-1}(V)$ is open in $X$. Since we know that every connected component of $X$ is open, the resulting topology is discrete.

(b) Let $\mathbb{T} \subseteq \mathbb{R}^2$ be the 'topologist's sine curve' defined as follows:

$$\mathbb{T}=\{(0,0)\} \cup \left\{(x,y) : y=\sin \frac{1}{x},\; 0 < x \leqslant 1 \right\}.$$

It follows that the quotient topology on the set of connected components of $\mathbb{T}$ is homeomorphic to the Sierpinski space $\mathbb{S}=\{0,1\}$ with the topology $\{\emptyset,\{1\},\{0,1\}\}$, which is not discrete.

4. (a) Show that if $X$ is a compact, Hausdorff topological space that contains more than one point, then there is a non-constant continuous function $f : X \to \mathbb{R}$.

   (b) Let $(X,d)$ be a compact metric space, let $a \in X$ and let $\mathcal{F}$ denote the set of functions $f : X \to \mathbb{R}$ such that $|f(x)| \leqslant d(x,a)$ for all $x \in X$. Show that $\mathcal{F}$ is compact in the topology of pointwise convergence.

   (c) Show that if $X$ is a locally compact, Hausdorff, $\sigma$-compact, second-countable topological space, then $X$ is completely metrizable.

(a) Let $X$ be a compact, Hausdorff topological space that contains more than one point. Since $X$ is compact and Hausdorff, $X$ is normal³. Thus, if we denote two disjoint closed subsets of $X$ by $A$ and $B$ (which is possible since $X$ has more than one point), by the Urysohn lemma, there exists a continuous map $f : X \to [0,1]$ where $f(x)=0$ for every $x \in A$ and $f(x)=1$ for every $x \in B$. By 'embedding' the function into $\mathbb{R}$, we obtain a non-constant, continuous function $f : X \to \mathbb{R}$ as desired.

(b) Note that the topology of pointwise convergence is equivalent to the product topology. Let $a \in X$ be arbitrarily chosen and fixed. We can then view $\mathcal{F}$ as a subset of $\mathbb{R}^X$ such that each of its factors $F \subseteq \mathbb{R}$ is bounded by a constant $d(x,a)$ in a weak inequality for each $x \in X$. This implies that $F$ is closed and bounded, and thus compact. By Tychonoff theorem, it follows that $\mathcal{F} \subseteq \mathbb{R}^X$ is compact in the topology of pointwise convergence.

(c) In fact, if a space $X$ is locally compact and second-countable, it is $\sigma$-compact⁴.

Since every locally compact, Hausdorff space is regular⁵, combined with Urysohn metrization theorem, $X$ is metrizable. Let $\tilde{X}$ be the completion of $X$⁶. We claim that $X$ is open in $\tilde{X}$ and thus completely metrizable⁴. To see this, let $h : X \to \tilde{X}$ be an isometric embedding. Let $x \in X$ and $V$ be a neighbourhood of $x$ that is contained in a compact subspace $C$ of $X$. It follows that $h(C)$ is a compact (also equivalently, sequentially compact) subset of $\tilde{X}$ so it cannot contain any element $y \in \tilde{X} \backslash h(X)$. By the same argument, $C=h(C)$, so $x \in h(C)$ and $h(C) \subseteq h(X)$, which implies that for every $x$ in $h(X)$, there exists a neighbourhood of $x$ with respect to $\tilde{X}$ that is contained in $h(X)$, so $h(X)$ is open in $\tilde{X}$.

5. Let $(Y,d)$ be a metric space, and let $(f_n)_{n=1}^{\infty}$ be a sequence of continuous maps from $\mathbb{R}^n$ to $Y$. Denote $\mathcal{F} := \{ f_n : n \in \mathbb{Z}^+ \}$.

   (a) Show that the compact-open topology on the space $\mathcal{C}(\mathbb{R}^n,Y)$ is Hausdorff.

   (b) Show that if $(f_n)_{n=1}^{\infty}$ converges pointwise to map $f : \mathbb{R}^n \to Y$ and $\mathcal{F}$ is an equicontinuous family, then $f$ is continuous.

   (c) Show that if $(f_n)_{n=1}^{\infty}$ converges in the compact open topology to a continuous map $f : \mathbb{R}^n \to Y$, then $\mathcal{F}$ is an equicontinuous family.

(a) We want to prove that if $X$ and $Y$ are topological spaces, then in the compact-open topology, $\mathcal{C}(X,Y)$ is Hausdorff if $Y$ is Hausdorff.⁷

Recall the definition of a subbasis element of a compact-open topology: if $X$ and $Y$ are topological spaces, $C$ is a compact subspace of $X$ and $W$ is an open subset of $Y$, then a subbasis element $S(C,W)$ of the compact-open topology is defined as follows:

$$S(C,W)=\{ f : f \in \mathcal{C}(X,Y) \text{ and } f(C) \subseteq W\}.$$

Let $f_1,f_2$ be two distinct elements in $\mathcal{C}(X,Y)$, then there exists an element $x_0 \in X$ such that $f_1(x_0) \neq f_2(x_0)$. Since $Y$ is Hausdorff, we obtain a neighbourhood $U$ of $f_1(x_0) \in Y$ and $V$ of $f_2(x_0) \in Y$ such that $U \cap V = \emptyset$. Pick $C=\{x_0\}$; it is compact since it is finite. It follows by the definition of the subbasis element above that $S(C,U)$ and $S(C,V)$ are disjoint neighbourhoods of $f_1$ and $f_2$ respectively. Thus, the compact-open topology is Hausdorff.

Finally, since $Y$ is a metric space, $Y$ is also Hausdorff, thus from the result above the compact-open topology on $\mathcal{C}(\mathbb{R}^n,Y)$ is Hausdorff, as desired.

(b) Let $(f_n)_{n=1}^{\infty}$ be a sequence of continuous maps from $\mathbb{R}^n$ to $Y$ that converges pointwise to a function $f : \mathbb{R}^n \to Y$. Note that $\mathbb{R}^n$ is metrizable, so we can use the sequential criterion of continuity to prove the continuity of $f$.

Let $\varepsilon>0$ be arbitrary. Let $(x_k)$ be a sequence in $\mathbb{R}^n$ that converges to a point $x \in \mathbb{R}^n$, then for each neighbourhood $U$ of $x$, there exists a positive integer $K_U$ such that $x_k \in U$ whenever $k \geqslant K_U$.

Let $f(x_k)$ be a sequence in $Y$. Due to the pointwise convergence of $(f_n)_{n=1}^{\infty}$ to $f$, there exist positive integers $N_{x_k}$ for each $k \in \mathbb{Z}_+$, as well as a positive integer $N_{x}$, such that $d(f_n(x_k),f(x_k))<\varepsilon/3$ and $d(f_n(x),f(x))<\varepsilon/3$ whenever $n \geqslant N_{x_k}$ and $n \geqslant N_x$ respectively. Due to the equicontinuity of $\mathcal{F}$, in particular at the limit point $x$ of the sequence $(x_k)$, there exists a neighbourhood $U'$ of $x$ such that for all $x' \in U'$ and $n \in \mathbb{Z}_+$, we have $d(f_n(x'),f_n(x))<\varepsilon/3$.

Now, choose a positive integer $K$ such that $K$ is larger than $K_{U'}$, and a positive integer $N$ such that $N$ is larger than $N_x$ and $N_{x_k}$ for each (not all, each) $k$. It follows by triangle inequality that whenever $n \geqslant N$ and $k \geqslant K$, we have that

$$d(f(x_k),f(x)) \leqslant d(f(x_k),f_n(x_k))+d(f_n(x_k),f_n(x))+d(f_n(x),f(x))<\varepsilon.$$

This implies that the sequence $f(x_k)$ converges to $f(x)$, so $f$ is continuous.

(c) Suppose that $(f_n)_{n=1}^{\infty}$ converges in the compact open topology to a continuous map $f : \mathbb{R}^n \to Y$. Note that since $Y$ is a metric space, the compact open topology is equal to the topology of compact convergence⁸. This implies that for each compact subspace $C$ of $\mathbb{R}^n$, the sequence $f_n|_{C}$ converges uniformly to $f|_{C}$⁹.

Let $\varepsilon>0$ be arbitrary. Let $x_0 \in \mathbb{R}^n$. Utilising its local compactness, let $U$ be a neighbourhood of $x_0$ that is contained in some compact subspace $C$. Due to the aforementioned uniform convergence, there exists a positive integer $N$ such that for each $x' \in U \subseteq C$, we have that $d(f_n|_{C}(x'),f|_{C}(x'))<\varepsilon/3$. Due to the continuity of the limit function $f$, there exists a neighbourhood $U'$ of $x_0$ such that for each $x'' \in U'$, we have that $d(f(x''),f(x_0))<\varepsilon/3$.

Applying triangle inequality, it follows that whenever $x \in U \cap U'$ (which is also a neighbourhood of $x_0$ that is contained in $C$) and $n \geqslant N$, we have that

$$d(f_n(x),f_n(x_0)) \leqslant d(f_n(x),f(x))+d(f(x),f(x_0))+d(f(x_0),f_n(x_0))<\varepsilon.$$

As for the terms in the sequence $(f_n)_{n=1}^{\infty}$ where $n < N$, utilising the continuity of each $f_n$, choose a neighbourhood $U_n$ of $x_0$ such that $d(f_n(x),f_n(x_0))<\varepsilon$ for each $x \in U_n$. It follows that for the neighbourhood $U_1 \cap \cdots \cap U_{N-1}$ of $x_0$, we have $d(f_n(x),f_n(x_0))<\varepsilon$ for all $n < N$.

Combining the two cases together, we obtain a neighbourhood of $x_0$, specifically $U_1 \cap \cdots \cap U_{N-1} \cap U \cap U'$, such that $d(f_n(x),f_n(x_0))<\varepsilon$ for all points $x$ in the neighbourhood and all $n \in \mathbb{Z}_+$. Therefore, $\mathcal{F}$ is equicontinuous.

Footnotes

1. See Theorem 25.4 in Chapter 3 of Munkres. ↩

2. See Theorem 25.5 in Chapter 3 of Munkres. ↩

3. See Theorem 32.3 in Chapter 4 of Munkres. ↩

4. See Exercise 46.10 in Chapter 7 of Munkres. ↩ ↩²

5. See Exercise 32.3 in Chapter 4 of Munkres. ↩

6. This is possible due to Theorem 43.7 in Chapter 7 of Munkres. ↩

7. See also Exercise 46.6 in Chapter 7 of Munkres. ↩

8. See Theorem 46.8 in Chapter 7 of Munkres. ↩

9. See Theorem 46.2 in Chapter 7 of Munkres. ↩