Proof Study Sheet

MT434P Topology · Midterm · Mon 30 Mar 2026
Exam format: one short proof covered in class — "something significant, short, key idea, should be possible to construct from scratch." Below are the 10 strongest candidates from your notes, ranked by likelihood. Each entry gives the statement, the key idea to remember, and the full proof.
Tier 1 — Very likely
Thm 2.1.8 HIGH

Closed ⟺ contains all its limit points

$F \subseteq (X,T)$ is closed $\iff$ $F = \overline{F}$ (i.e. $F$ contains all its limit points).
Key idea: (⇒) If $x \notin F$, then $X \setminus F$ is an open set separating $x$ from $F$, so $x$ is not a limit point. (⇐) Every point of $X \setminus F$ has an open neighbourhood missing $F$, so $X \setminus F$ is a neighbourhood of each of its points, hence open by Lemma 2.1.3.
(⇒)

$F$ closed, so $X \setminus F$ is open. Let $x \notin F$. Then $x \in X \setminus F$ and $(X \setminus F) \cap F = \emptyset$, so $(X \setminus F) \cap F \setminus \{x\} = \emptyset$. Hence $x$ is not a limit point. Therefore $F$ contains all its limit points.

(⇐)

Assume $\overline{F} = F$. Take any $y \in X \setminus F$. Since $y$ is not a limit point of $F$, $\exists\, U$ open with $y \in U$ and $U \cap F = \emptyset$, so $y \in U \subseteq X \setminus F$. Thus $X \setminus F$ is a neighbourhood of each of its points, so by Lemma 2.1.3, $X \setminus F$ is open, and $F$ is closed.

∎
Uses: Lemma 2.1.3
Lemma 2.1.3 HIGH

Open ⟺ neighbourhood of each of its points

$U \subseteq (X,T)$ is open $\iff$ $U$ is a neighbourhood of each of its points.
Key idea: (⇒) trivial. (⇐) For each $x \in U$ pick an open $V_x \subseteq U$. Then $U = \bigcup_{x \in U} V_x$, a union of open sets.
(⇒)

$U$ open, so $U$ is trivially a neighbourhood of each $x \in U$ (take $V = U$).

(⇐)

For each $x \in U$, since $U$ is a neighbourhood of $x$, $\exists\, V_x \in T$ with $x \in V_x \subseteq U$. Then $\bigcup_{x \in U} V_x \subseteq U$. But $x \in V_x$ for each $x$, so $U \subseteq \bigcup_{x \in U} V_x$. Therefore $U = \bigcup_{x \in U} V_x$, a union of open sets, hence open.

∎
Thm 2.1.5 HIGH

Closed sets: arbitrary intersections & finite unions

In any topological space: (1) arbitrary intersections of closed sets are closed, (2) finite unions of closed sets are closed.
Key idea: Take complements and apply De Morgan's laws to reduce to the open-set axioms.
(1)

Let $\{F_\alpha\}_{\alpha \in \Lambda}$ be closed. Then $X \setminus \bigcap_\alpha F_\alpha = \bigcup_\alpha (X \setminus F_\alpha)$. Each $X \setminus F_\alpha$ is open, so the RHS is a union of open sets, hence open. Therefore $\bigcap_\alpha F_\alpha$ is closed.

(2)

Let $F_1, \dots, F_n$ be closed. Then $X \setminus \bigcup_{i=1}^n F_i = \bigcap_{i=1}^n (X \setminus F_i)$. Each $X \setminus F_i$ is open, so the RHS is a finite intersection of open sets, hence open. Therefore $\bigcup_{i=1}^n F_i$ is closed.

∎
Thm 2.3.2 HIGH

Continuity can be checked on a basis

Let $\mathcal{B}$ be a basis for $T'$. Then $f:(X,T) \to (Y,T')$ is continuous $\iff$ $f^{-1}(B) \in T$ for every $B \in \mathcal{B}$.
Key idea: (⇒) trivial since $\mathcal{B} \subseteq T'$. (⇐) Any open $V$ is a union of basis sets; preimage of a union is the union of preimages; union of open sets is open.
(⇒)

$\mathcal{B} \subseteq T'$, so preimages of basis sets are preimages of open sets, hence open.

(⇐)

Let $V \in T'$. By Thm 2.2.3, $V = \bigcup_\alpha B_\alpha$ for some $\{B_\alpha\} \subseteq \mathcal{B}$. Then $f^{-1}(V) = f^{-1}(\bigcup_\alpha B_\alpha) = \bigcup_\alpha f^{-1}(B_\alpha)$. Each $f^{-1}(B_\alpha)$ is open by assumption, so $f^{-1}(V)$ is a union of open sets, hence open.

∎
Uses: Thm 2.2.3
Thm 2.3.3 HIGH

Continuous ⟺ preimages of closed sets are closed

$f:(X,T) \to (Y,T')$ is continuous $\iff$ $f^{-1}(C)$ is closed in $X$ for every closed $C \subseteq Y$.
Key idea: Use $f^{-1}(Y \setminus C) = X \setminus f^{-1}(C)$ to flip between open and closed preimages.
(⇒)

$C$ closed in $Y$, so $Y \setminus C$ is open. By continuity $f^{-1}(Y \setminus C)$ is open in $X$. Since $f^{-1}(Y \setminus C) = X \setminus f^{-1}(C)$, we get $f^{-1}(C)$ is closed.

(⇐)

Let $V$ be open in $Y$, so $Y \setminus V$ is closed. By assumption $f^{-1}(Y \setminus V)$ is closed, so $X \setminus f^{-1}(Y \setminus V) = f^{-1}(V)$ is open.

∎
Tier 2 — Likely
Thm 2.1.12 MED

Interior = union of all open subsets

For $A \subseteq (X,T)$: $\mathrm{int}(A)$ equals the union of all open subsets of $A$, and so is open.
Key idea: Show double containment. If $a \in \mathrm{int}(A)$, its witnessing open set lies in $V$. If $v \in V$, some open subset of $A$ contains $v$, so $v \in \mathrm{int}(A)$.

Let $V = \bigcup \{U \in T \mid U \subseteq A\}$.

If $a \in \mathrm{int}(A)$, then $\exists\, U \in T$ with $a \in U \subseteq A$, so $U \subseteq V$, hence $a \in V$. Thus $\mathrm{int}(A) \subseteq V$.

If $v \in V$, then $v \in U$ for some $U \in T$ with $U \subseteq A$. By definition $v \in \mathrm{int}(A)$. Thus $V \subseteq \mathrm{int}(A)$.

∎
Cor 2.1.10 MED

Closure = smallest closed superset

$\overline{A}$ is the smallest closed set containing $A$, and equals the intersection of all closed supersets of $A$.
Key idea: If $B$ is closed with $A \subseteq B \subseteq \overline{A}$, then every limit point of $A$ is a limit point of $B$, and $B$ is closed so contains them. Thus $B \supseteq \overline{A}$, forcing $B = \overline{A}$.

Suppose $B$ is closed with $A \subseteq B \subseteq \overline{A}$. Every limit point of $A$ is a limit point of $B$ (since $A \subseteq B$ means $U \cap A \setminus \{x\} \neq \emptyset \Rightarrow U \cap B \setminus \{x\} \neq \emptyset$). Since $B$ is closed, it contains all its limit points (Thm 2.1.8). So $B \supseteq A \cup \{\text{limit points of } A\} = \overline{A}$. Hence $B = \overline{A}$.

∎
Uses: Thm 2.1.8
Thm 2.1.17 MED

Closure = A ∪ ∂A

$\overline{A} = A \cup \partial A$ where $\partial A = \overline{A} \cap \overline{X \setminus A}$.
Key idea: ($\subseteq$) trivially both $A$ and $\partial A$ sit inside $\overline{A}$. ($\supseteq$) if $x \in \overline{A}$ and $x \notin A$, then $x \in X \setminus A \subseteq \overline{X \setminus A}$, so $x \in \partial A$.

$A \subseteq \overline{A}$ and $\partial A = \overline{A} \cap \overline{X \setminus A} \subseteq \overline{A}$, so $A \cup \partial A \subseteq \overline{A}$.

Conversely, let $x \in \overline{A}$. If $x \in A$, then $x \in A \cup \partial A$. If $x \notin A$, then $x \in X \setminus A \subseteq \overline{X \setminus A}$, so $x \in \overline{A} \cap \overline{X \setminus A} = \partial A \subseteq A \cup \partial A$.

∎
Thm 3.1.6 MED

Restriction of a continuous map is continuous

If $f:(X,T) \to (Y,T')$ is continuous and $A \subseteq X$, then $f|_A : (A, T_A) \to (Y, T')$ is continuous.
Key idea: $(f|_A)^{-1}(V) = A \cap f^{-1}(V)$. Since $f^{-1}(V)$ is open in $X$, the intersection with $A$ is open in $T_A$ by definition of subspace topology.

Let $V$ be open in $(Y, T')$. By continuity, $f^{-1}(V)$ is open in $(X,T)$. Now $(f|_A)^{-1}(V) = A \cap f^{-1}(V)$, which is open in $(A, T_A)$ by definition of the subspace topology.

∎
Lemma 3.1.5 MED

Closed in subspace ⟺ intersection with closed set

$F \subseteq A$ is closed in $(A, T_A)$ $\iff$ $F = A \cap F'$ for some $F'$ closed in $(X, T)$.
Key idea: Mirror the subspace topology definition ($T_A$ intersects open sets with $A$) but for closed sets. Use complements: closed in $A$ means complement open in $A$, which is $A \cap U$ for some $U$ open in $X$, so $F = A \setminus (A \cap U) = A \cap (X \setminus U)$.
(⇒)

$F$ closed in $(A, T_A)$ means $A \setminus F \in T_A$, so $\exists\, U \in T$ with $A \cap U = A \setminus F$. Set $F' = X \setminus U$ (closed in $X$). Then $F' \cap A = (X \setminus U) \cap A = A \setminus (U \cap A) = A \setminus (A \setminus F) = F$.

(⇐)

$F = A \cap F'$ with $F'$ closed, so $F' = X \setminus U$ for some $U \in T$. Then $A \setminus F = A \setminus (A \cap F') = A \cap (X \setminus F') = A \cap U \in T_A$. So $F$ is closed in $T_A$.

∎