What Is the Relationship Between Compact and Hausdorff Spaces?

“Compact space and Hausdorff space” can mean either a comparison of the two properties or a question about what they imply together. The short answer is that they are independent: a space can have either property without the other. When both properties appear in a theorem, however, they often give strong conclusions about closed sets and continuous maps.

The two definitions

A topological space $X$ is compact if every open cover of $X$ has a finite subcover. In other words, if open sets cover all of $X$, finitely many of them already cover all of $X$. This definition does not require $X$ to be Hausdorff.

A space $X$ is Hausdorff if any two distinct points have disjoint open neighborhoods. For every $x \ne y$, there must be open sets $U$ and $V$ with $x \in U$, $y \in V$, and $U \cap V = \varnothing$.

These conditions address different features. Compactness limits how open covers can behave; the Hausdorff condition ensures distinct points can be separated by neighborhoods.

Neither property implies the other

An infinite set with the indiscrete topology—whose only open sets are the empty set and the whole space—is compact. Any open cover of a nonempty such space must include the whole space, so a single set from the cover suffices. But this space is not Hausdorff: there are no two disjoint nonempty open neighborhoods of distinct points.

For the reverse direction, the real line with its usual topology is Hausdorff but not compact. For example, the open sets $(-n,n)$ for positive integers $n$ cover $\mathbb{R}$, but no finite selection covers the whole line: the largest chosen interval still has bounded endpoints.

The usual interval $[0,1]$ is both compact and Hausdorff. This familiar example is useful, but it should not suggest that the two properties are equivalent or that either one follows from the other.

Why compact subsets of Hausdorff spaces are closed

Let $K$ be a compact subset of a Hausdorff space $X$. To show that $K$ is closed, take any point $x \in X \setminus K$. For every $y \in K$, the Hausdorff property gives disjoint open neighborhoods $U_y$ of $x$ and $V_y$ of $y$.

The sets $V_y$ cover $K$. Since $K$ is compact, finitely many cover it, say $V_{y_1}, \ldots, V_{y_m}$. The intersection

$$U = U_{y_1} \cap \cdots \cap U_{y_m}$$

is an open neighborhood of $x$. It is disjoint from every selected $V_{y_i}$, and those sets cover $K$, so $U$ does not meet $K$. Thus every point outside $K$ has an open neighborhood outside $K$. The complement of $K$ is open, which means $K$ is closed.

This theorem is often written as “compact subsets of Hausdorff spaces are closed.” It does not say that every closed subset of every space is compact. A closed subset of a compact space is compact, but that is a different result and does not require the space to be Hausdorff.

A useful theorem about continuous bijections

Suppose $f:X \to Y$ is a continuous bijection, $X$ is compact, and $Y$ is Hausdorff. Then $f$ is a homeomorphism: its inverse is continuous as well.

Here is the reason. A closed subset $C$ of compact $X$ is compact, so its continuous image $f(C)$ is compact in $Y$. Since $Y$ is Hausdorff, compact subsets of $Y$ are closed. Therefore $f$ maps closed sets to closed sets. A bijection that maps closed sets to closed sets has a continuous inverse.

For a concrete application, $f:[0,1] \to [0,1]$ given by $f(x)=x^3$ is a continuous bijection. The interval is compact and Hausdorff, so the theorem guarantees that its inverse is continuous; indeed, $f^{-1}(y)=\sqrt[3]{y}$. The theorem is more useful when the inverse is difficult to check directly.

Both hypotheses play a role. For example, the map $t \mapsto (\cos t,\sin t)$ from $[0,2\pi)$ to the unit circle is a continuous bijection, but its inverse is not continuous at $(1,0)$. The domain is not compact, so the theorem does not apply.

Common mistakes

• Assuming compact means compact Hausdorff. Compactness alone does not include a separation condition. Some authors use “compact” to mean “compact Hausdorff” in particular contexts, so check the stated convention.
• Reversing the compact-subset theorem. In a Hausdorff space, compact subsets are closed. A closed subset need not be compact unless additional conditions—such as compactness of the whole space—are available.
• Requiring the domain to be Hausdorff in the bijection theorem. The theorem needs a compact domain and a Hausdorff codomain. It does not require the domain to be Hausdorff.
• Treating the real-line example as a proof that Hausdorff spaces are never compact. The line is one example; many spaces, including $[0,1]$, are both compact and Hausdorff.