Common Pitfalls

1. “Arbitrary intersections of open sets are open.” False. Only finite intersections are guaranteed. Counterexample: in $\mathbb{R}$, $\bigcap_{n=1}^\infty (-1/n, 1/n) = \{0\}$, which is not open.

2. “Closed = not open.” False. A set can be both (clopen), neither, or exactly one. In $\mathbb{R}$, $\emptyset$ and $\mathbb{R}$ are clopen; $[0, 1)$ is neither.

3. “Compact implies closed and bounded in every topological space.” False. This is specific to $\mathbb{R}^n$ (Heine—Borel). In the cofinite topology on an infinite set, every subset is compact but not every subset is closed.

4. “Connected implies path-connected.” False. The topologist’s sine curve is connected but not path-connected.

5. “Continuous bijections are homeomorphisms.” False. The bijection $f : [0, 2\pi) \to S^1$ given by $f(t) = (\cos t, \sin t)$ is continuous and bijective, but its inverse is not continuous — $[0, 2\pi)$ is not compact but $S^1$ is.

6. “Every metric space is complete.” False. $\mathbb{Q}$ with the usual metric is not complete.

7. “The closure of the interior equals the interior of the closure.” False as a general principle. In $\mathbb{R}$ with $A = \mathbb{Q} \cap (0, 1)$: $\operatorname{int}(\overline{A}) = (0, 1)$ but $\overline{\operatorname{int}(A)} = \emptyset$.

8. “A compact subset of a Hausdorff space is closed.” This is true. But the converse requires the space to be compact: a closed subset of a compact space is compact. These are dual statements that are frequently confused.

9. “Every continuous function is uniformly continuous on a compact set.” True for metric spaces, but the concept of uniform continuity requires a metric. In a general topological space, uniform continuity is not defined. The correct statement: a continuous function on a compact metric space is uniformly continuous (Heine-Cantor theorem).

10. “Product of connected sets is connected.” True for finite products. For infinite products, the product of connected spaces is connected in the product topology (true), but this requires the product topology, not the box topology. The box product of connected spaces can be disconnected.

11. “Quotient maps are open.” False. The quotient map $q : X \to X/{\sim}$ is not necessarily open. For example, collapsing the boundary of a disk to a point yields a sphere, but the image of a small open set at the boundary may not be open.

12. “A subspace of a compact space is compact.” False. Only closed subspaces of compact spaces are compact. The open interval $(0,1)$ is not compact as a subspace of the compact space $[0,1]$.

13. “Every limit point is a boundary point.” False. Interior points can also be limit points. In $\mathbb{R}$, $0$ is a limit point of $(-1, 1)$ but is not a boundary point ($\partial(-1,1) = \{-1, 1\}$).

14. “If a space is separable, it is second countable.” False. $\mathbb{R}$ with the lower-limit topology (Sorgenfrey line) is separable ($\mathbb{Q}$ is dense) but not second countable.

“Compact implies closed in a Hausdorff space.” Let $K \subseteq X$ be compact and $X$ Hausdorff. For any $x \in X \setminus K$, for each $y \in K$, choose disjoint open $U_y \ni x$, $V_y \ni y$. Cover $K$ by finitely many $V_{y_i}$; let $U = \bigcap U_{y_i}$. Then $U$ is an open neighborhood of $x$ disjoint from $K$, so $X \setminus K$ is open. $\blacksquare$

“Closed subset of a compact space is compact.” Let $K \subseteq X$ be closed, $X$ compact, and $\{U_\alpha\}$ an open cover of $K$. Add $X \setminus K$ to get an open cover of $X$, extract a finite subcover, and remove $X \setminus K$ if included. $\blacksquare$

“Continuous image of a compact set is compact.” Let $f : X \to Y$ be continuous, $K \subseteq X$ compact. For any open cover $\{V_\alpha\}$ of $f(K)$, $\{f^{-1}(V_\alpha)\}$ is an open cover of $K$. Extract a finite subcover; the corresponding $V_\alpha$ cover $f(K)$. $\blacksquare$

Problem 1. Show that $X = \mathbb{N}$ with the cofinite topology is compact but not Hausdorff.

Solution. For compactness: let $\{U_\alpha\}$ be any open cover. Pick any non-empty $U_{\alpha_0}$. Since only finitely many points are missing from $U_{\alpha_0}$, pick one $U_\alpha$ for each missing point. This gives a finite subcover. For Hausdorff: any two non-empty open sets in the cofinite topology intersect (since their complements are finite), so $X$ is not Hausdorff. $\blacksquare$

Problem 2. Prove that the topologist’s sine curve is connected but not path-connected.

Solution. Let $S = \{(x, \sin(1/x)) : x > 0\} \cup \{(0, y) : |y| \leq 1\}$. $S$ is connected because the graph of $\sin(1/x)$ is connected and its closure adds $\{0\} \times [-1,1]$. For path-connectedness: suppose a path $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ connects $(0,0)$ to $(1/\pi, 0)$. Let $t_0 = \inf\{t : \gamma_1(t) > 0\}$. By continuity, $\gamma_1(t_0) = 0$. For any $\delta > 0$, the image of $(t_0, t_0 + \delta)$ is connected but the graph oscillates infinitely often, contradicting continuity. $\blacksquare$

• Are you assuming a metric space? Many properties (uniform continuity, completeness, sequential compactness) only make sense in metric spaces.
• Does the statement involve “closed and bounded”? That equivalence only holds in $\mathbb{R}^n$.
• Are you confusing sequentially compact with compact? They are equivalent in metric spaces but not in general topological spaces.
• Is the space Hausdorff? Many theorems about compactness and uniqueness of limits require Hausdorff separation.

15. “The product of Hausdorff spaces is Hausdorff.” This is true for the product topology but false for the box topology on an infinite product. In the box topology, the product of Hausdorff spaces may fail to be Hausdorff because the basis elements are too restrictive.

16. “Every continuous function on a compact set attains its maximum.” True for functions into $\mathbb{R}$ (extreme value theorem), but the codomain matters. A continuous function from a compact space into an arbitrary topological space need not attain a “maximum” — the concept of maximum requires an order structure.

17. “A subspace of a connected space is connected.” False. The interval $(0,1) \cup (2,3)$ is a subspace of the connected space $\mathbb{R}$ but is disconnected. Connectedness is not hereditary; only open connected subspaces inherit connectedness in general.

18. “All open covers of a compact space have a finite subcover.” This is the definition of compactness, so it is true by definition. However, a common mistake is thinking that “every open cover has a finite subcover” is a property to be proven rather than the definition. The difficulty lies in finding the finite subcover, not in stating the definition.

19. “If $X$ is compact and $f : X \to Y$ is continuous, then $f$ is a homeomorphism onto its image if $f$ is injective.” This requires $Y$ to be Hausdorff. Counterexample: the identity map from $[0,1]$ with the discrete topology to $[0,1]$ with the standard topology is continuous and bijective but not a homeomorphism (the domain is not compact in the discrete topology). For a correct statement: a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

20. “The one-point compactification of a space is always Hausdorff.” False. The one-point compactification $X^* = X \cup \{\infty\}$ is Hausdorff if and only if $X$ is locally compact and Hausdorff. For example, the one-point compactification of $\mathbb{Q}$ is not Hausdorff because $\mathbb{Q}$ is not locally compact.