Topology Syllabus

State four equivalent characterizations of continuity of $f : X \to Y$.

$f$ is continuous iff: (1) $f^{-1}(V)$ open for every open $V \subseteq Y$; (2) $f^{-1}(C)$ closed for every closed $C \subseteq Y$; (3) $f(\overline{A}) \subseteq \overline{f(A)}$ for all $A \subseteq X$; (4) for each $x$ and each neighborhood $V$ of $f(x)$, there is a neighborhood $U$ of $x$ with $f(U) \subseteq V$.

State the pasting lemma for continuous functions.

Let $X = A \cup B$ where $A, B$ are both closed (or both open) in $X$. If $f : A \to Y$ and $g : B \to Y$ are continuous and $f(x) = g(x)$ for all $x \in A \cap B$, then the combined map $h : X \to Y$ defined by $f$ on $A$ and $g$ on $B$ is continuous.

Define a homeomorphism $f : X \to Y$.

A homeomorphism is a bijection $f : X \to Y$ such that both $f$ and its inverse $f^{-1}$ are continuous. Equivalently, $f$ is a continuous bijection that is also an open map (maps open sets to open sets). It establishes a one-to-one correspondence of open sets.

Define a topological property and give three examples.

A topological property is one preserved by homeomorphisms (invariant under the equivalence). Examples: compactness, connectedness, the Hausdorff ($T_2$) property, cardinality, and separability. Two homeomorphic spaces share all topological properties.

Distinguish a continuous bijection from a homeomorphism with an example.

A continuous bijection need not be a homeomorphism because $f^{-1}$ may fail to be continuous. Example: $f : [0,1) \to S^1$, $f(t) = (\cos 2\pi t, \sin 2\pi t)$, is a continuous bijection but not a homeomorphism since $f^{-1}$ is discontinuous at $(1,0)$.

Define an embedding (topological imbedding) $f : X \to Y$.

$f : X \to Y$ is an embedding if $f$ is injective, continuous, and the corestriction $f : X \to f(X)$ is a homeomorphism onto its image $f(X)$ with the subspace topology. Thus $X$ is realized as a subspace of $Y$.

In a topological space, what does it mean for a sequence $(x_n)$ to converge to $x$?

$(x_n) \to x$ if for every open set (neighborhood) $U$ containing $x$, there exists $N \in \mathbb{N}$ such that $x_n \in U$ for all $n \geq N$. That is, the sequence is eventually inside every neighborhood of $x$.

Why can sequences converge to more than one limit, and what condition prevents this?

In a general topological space limits need not be unique (e.g. in the indiscrete topology every sequence converges to every point). In a Hausdorff ($T_2$) space limits of sequences are unique, since distinct points have disjoint neighborhoods.

State the sequence lemma relating convergence and closure in a metric space.

In a metric space (more generally, a first-countable space), $x \in \overline{A}$ iff there is a sequence of points of $A$ converging to $x$. Also $f$ is continuous iff $x_n \to x$ implies $f(x_n) \to f(x)$ (sequential continuity equals continuity).

Define a limit point (accumulation point) of a subset $A$ of a topological space $X$.

A point $x \in X$ is a limit point of $A$ if every open set $U$ containing $x$ intersects $A$ in some point other than $x$, i.e. $(U \setminus \{x\}) \cap A \neq \varnothing$. The set of limit points is denoted $A'$ (the derived set).

State the formula relating the closure of $A$ to $A$ and its set of limit points $A'$.

$$\overline{A} = A \cup A'.$$ Consequently, $A$ is closed iff $A' \subseteq A$, i.e. iff $A$ contains all of its limit points.

Define an isolated point of a set $A$.

A point $x \in A$ is an isolated point of $A$ if it is not a limit point of $A$, i.e. there exists an open set $U$ with $U \cap A = \{x\}$. In a discrete space every point is isolated and $A' = \varnothing$ for every $A$.

Define a closed set and give three equivalent conditions for $A$ to be closed.

$A \subseteq X$ is closed if its complement $X \setminus A$ is open. Equivalently: (1) $A = \overline{A}$; (2) $A' \subseteq A$ (contains all limit points); (3) the complement is open. Arbitrary intersections and finite unions of closed sets are closed.

Define the closure $\overline{A}$ of a set $A$ and describe it as the smallest closed set.

The closure $\overline{A}$ is the intersection of all closed sets containing $A$; it is the smallest closed set containing $A$. A point $x \in \overline{A}$ iff every neighborhood of $x$ meets $A$.

Define the interior $\operatorname{Int}(A)$ of a set $A$ and describe it as the largest open set.

The interior $\operatorname{Int}(A)$ is the union of all open sets contained in $A$; it is the largest open set inside $A$. A point $x \in \operatorname{Int}(A)$ iff some open set $U$ satisfies $x \in U \subseteq A$.

State the duality relating interior and closure via complements.

$$\operatorname{Int}(A) = X \setminus \overline{X \setminus A}, \qquad \overline{A} = X \setminus \operatorname{Int}(X \setminus A).$$ Interior of $A$ is the complement of the closure of the complement, and vice versa.

Define the boundary $\partial A$ of a set $A$ and express it using closure.

$$\partial A = \overline{A} \cap \overline{X \setminus A} = \overline{A} \setminus \operatorname{Int}(A).$$ A point is in the boundary iff every neighborhood meets both $A$ and its complement. Also $\overline{A} = \operatorname{Int}(A) \sqcup \partial A$.

Define a dense subset $A$ of a space $X$ and give an example.

$A$ is dense in $X$ if $\overline{A} = X$, equivalently every nonempty open set of $X$ meets $A$. Example: $\mathbb{Q}$ is dense in $\mathbb{R}$ with the standard topology. A space with a countable dense subset is called separable.

Define a $T_0$ (Kolmogorov) space.

A space $X$ is $T_0$ if for any two distinct points $x \neq y$, there exists an open set containing one of them but not the other. Distinct points are topologically distinguishable.

Define a $T_1$ space and give an equivalent characterization.

$X$ is $T_1$ if for every pair of distinct points $x \neq y$ there is an open set containing $x$ but not $y$ (and one containing $y$ but not $x$). Equivalently, $X$ is $T_1$ iff every singleton $\{x\}$ (hence every finite set) is closed.

Define a $T_2$ (Hausdorff) space.

$X$ is Hausdorff ($T_2$) if for any two distinct points $x \neq y$ there exist disjoint open sets $U, V$ with $x \in U$, $y \in V$. In a Hausdorff space, limits of sequences are unique and every finite point set is closed.

State the implication chain among the separation axioms $T_2, T_1, T_0$.

$$T_2 \implies T_1 \implies T_0.$$ Each is strictly stronger. Example: an infinite set with the cofinite topology is $T_1$ but not $T_2$ (Hausdorff).

In a $T_1$ space, characterize a limit point $x$ of $A$ in terms of how many points of $A$ neighborhoods contain.

If $X$ is a $T_1$ space, then $x$ is a limit point of $A$ iff every neighborhood of $x$ contains infinitely many points of $A$. (This fails in general non-$T_1$ spaces.)

Define a regular space (and the $T_3$ property).

A space $X$ (assumed $T_1$) is regular if for every point $x$ and every closed set $C$ not containing $x$, there exist disjoint open sets $U, V$ with $x \in U$ and $C \subseteq V$. A $T_1$ regular space is called $T_3$.

Define a normal space (and the $T_4$ property).

A space $X$ (assumed $T_1$) is normal if for every pair of disjoint closed sets $C, D$ there exist disjoint open sets $U, V$ with $C \subseteq U$ and $D \subseteq V$. A $T_1$ normal space is called $T_4$.

State the implication chain among normal, regular, and Hausdorff (with $T_1$ assumed).

$$\text{Normal }(T_4) \implies \text{Regular }(T_3) \implies \text{Hausdorff }(T_2) \implies T_1 \implies T_0.$$ Each implication is strict; the converses fail in general.

State the neighborhood ('shrinking') characterization of regularity.

A $T_1$ space $X$ is regular iff for every point $x$ and every open set $U$ containing $x$, there is an open set $V$ with $x \in V \subseteq \overline{V} \subseteq U$. Open neighborhoods can be shrunk with closure inside.

State the closed-neighborhood characterization of normality.

A $T_1$ space $X$ is normal iff for every closed set $C$ and every open set $U \supseteq C$, there is an open set $V$ with $C \subseteq V \subseteq \overline{V} \subseteq U$.

State two important classes of spaces that are always normal.

(1) Every metrizable space is normal (indeed metric spaces are $T_4$). (2) Every compact Hausdorff space is normal. Also, every regular space with a countable basis (second countable regular space) is normal.

State Urysohn's Lemma.

If $X$ is a normal space and $A, B$ are disjoint closed subsets, then there exists a continuous function $f : X \to [0,1]$ such that $f(x) = 0$ for all $x \in A$ and $f(x) = 1$ for all $x \in B$. Such $f$ is called a Urysohn function.

Why is the function in Urysohn's Lemma called a 'separating' function, and does $f=0$ exactly on $A$?

It separates the disjoint closed sets $A$ and $B$ by real values $0$ and $1$, giving a continuous 'measurement' of closeness. In general $f$ need not equal $0$ only on $A$; the condition is $f(A) \subseteq \{0\}$, $f(B) \subseteq \{1\}$. Exact preimages (perfect separation) require the stronger perfectly-normal condition.

State the Tietze Extension Theorem.

If $X$ is a normal space and $A \subseteq X$ is closed, then any continuous function $f : A \to \mathbb{R}$ (or into $[a,b]$) can be extended to a continuous function $g : X \to \mathbb{R}$ (respectively into $[a,b]$) with $g|_A = f$. Normality is both necessary and sufficient.

How are Urysohn's Lemma and the Tietze Extension Theorem logically related?

For a $T_1$ space, normality is equivalent to the conclusion of Urysohn's Lemma and equivalent to the Tietze extension property; the two theorems are equivalent characterizations of normality. Urysohn's Lemma is the key tool used to prove Tietze's theorem.

Define an open cover of a space $X$ and a subcover.

An open cover of $X$ is a collection $\mathcal{A} = \{U_\alpha\}$ of open sets with $X = \bigcup_\alpha U_\alpha$. A subcover is a subcollection of $\mathcal{A}$ that still covers $X$. A finite subcover uses only finitely many of the $U_\alpha$.

Define a compact topological space.

A space $X$ is compact if every open cover of $X$ has a finite subcover. That is, for any collection of open sets covering $X$, finitely many of them already cover $X$.

State the finite intersection property (FIP) characterization of compactness.

$X$ is compact iff for every collection $\{C_\alpha\}$ of closed sets with the finite intersection property (every finite subcollection has nonempty intersection), the total intersection $\bigcap_\alpha C_\alpha$ is nonempty.

State the key facts about compactness under subspaces, closed sets, and continuous images.

(1) A closed subset of a compact space is compact. (2) A compact subset of a Hausdorff space is closed. (3) The continuous image of a compact space is compact. Consequently a continuous real function on a compact space attains its max and min (extreme value theorem).

State the theorem giving conditions under which a continuous bijection is automatically a homeomorphism.

If $f : X \to Y$ is a continuous bijection, $X$ is compact, and $Y$ is Hausdorff, then $f$ is a homeomorphism. (Because $f$ maps closed/compact sets to compact/closed sets, so $f^{-1}$ is continuous.)

State the Heine-Borel Theorem.

A subset $A \subseteq \mathbb{R}^n$ (with the standard/Euclidean topology) is compact if and only if it is closed and bounded. Boundedness means $A$ lies in some ball of finite radius.

Does the Heine-Borel characterization ('closed and bounded $\Rightarrow$ compact') hold in every metric space?

No. In a general metric space, compact implies closed and bounded, but the converse fails. Example: the closed unit ball of an infinite-dimensional normed space is closed and bounded but not compact. The correct general criterion is complete and totally bounded.

State the tube lemma.

If $Y$ is compact and $X \times Y$ contains an open set $N$ containing the slice $\{x_0\} \times Y$, then $N$ contains a 'tube' $W \times Y$ for some open set $W \ni x_0$ in $X$. It is a key tool for proving products of compact spaces are compact.

State the theorem on compactness of finite products of compact spaces (and name its infinite generalization).

The product of finitely many compact spaces is compact (proved via the tube lemma). The generalization to arbitrary products is the Tychonoff Theorem: any product $\prod_\alpha X_\alpha$ of compact spaces is compact in the product topology.

Define a sequentially compact space.

A space $X$ is sequentially compact if every sequence $(x_n)$ in $X$ has a subsequence $(x_{n_k})$ that converges to a point of $X$.

Define a limit point compact space.

A space $X$ is limit point compact (also called Frechet or Bolzano-Weierstrass compact) if every infinite subset $A \subseteq X$ has at least one limit point in $X$.

State the relationships among compactness, sequential compactness, and limit point compactness in general spaces.

In a general topological space: compact $\implies$ limit point compact, and sequentially compact $\implies$ limit point compact. The three notions are NOT equivalent in general; none of the reverse implications holds without extra hypotheses.

For metric spaces, how are compactness, sequential compactness, and limit point compactness related?

For metric spaces (and metrizable spaces) all three coincide: $$\text{compact} \iff \text{sequentially compact} \iff \text{limit point compact}.$$ This equivalence is a central theorem of metric topology.

State the Bolzano-Weierstrass theorem for $\mathbb{R}^n$.

Every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence; equivalently, every bounded infinite subset of $\mathbb{R}^n$ has a limit point. It is the special case of sequential/limit-point compactness for closed bounded subsets of $\mathbb{R}^n$.

State the Lebesgue number lemma.

If $\mathcal{A}$ is an open cover of a compact metric space $(X,d)$, then there exists $\delta > 0$ (a Lebesgue number) such that every subset of $X$ with diameter less than $\delta$ is contained in some single member of $\mathcal{A}$.

State the theorem that continuous functions on compact metric spaces are uniformly continuous.

If $f : X \to Y$ is continuous and $X$ is a compact metric space (with $Y$ metric), then $f$ is uniformly continuous: for every $\varepsilon > 0$ there is $\delta > 0$ such that $d_X(a,b) < \delta \implies d_Y(f(a), f(b)) < \varepsilon$ for all $a,b \in X$.

Define a bounded subset of a metric space and note the difference between 'bounded' and 'totally bounded'.

$A$ is bounded if there is $M$ with $d(x,y) \leq M$ for all $x,y \in A$ (finite diameter). $A$ is totally bounded if for every $\varepsilon > 0$ it is covered by finitely many $\varepsilon$-balls. Totally bounded $\implies$ bounded, but not conversely; compactness $=$ complete $+$ totally bounded.

Give the classification: which is finer, and which points of the standard vs discrete topology on $\mathbb{R}$ are isolated?

The discrete topology is strictly finer than the standard topology on $\mathbb{R}$. In the discrete topology every point is isolated and no set has limit points; in the standard topology no single point is isolated in $\mathbb{R}$ and every point is a limit point of any interval containing it.

Compare 'closed' and 'compact': give the key implications in a general space versus a Hausdorff space.

In general: a closed subset of a compact space is compact, but a closed set need not be compact (e.g. $\mathbb{R}$ is closed in itself, not compact). In a Hausdorff space, every compact subset is closed. So in compact Hausdorff spaces, closed $\iff$ compact.