Compactness and Tychonoff

In one sentence

Compactness is still every open cover having a finite subcover, but outside metric spaces it no longer coincides with sequential compactness, and Tychonoff — arbitrary products of compact spaces are compact — is the theorem that makes the product topology worth having.

Why the exam asks it

Because the metric-space equivalences are so useful that they get quoted as definitions, and in a general space they are false in both directions. Recognising which theorem needs which hypothesis is the whole skill.

The idea

$K$ is compact if every open cover has a finite subcover. That definition is unchanged. What changes is what it implies.

What survives in any space

• Continuous images of compact sets are compact.
• Closed subsets of compact sets are compact.
• Finite unions of compact sets are compact.
• A continuous real function on a compact space is bounded and attains its bounds.
• Compact $+$ Hausdorff $\Rightarrow$ normal, so all of Urysohn applies.
• A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

What needs Hausdorff

Compact subsets are closed. In the cofinite topology on $\mathbb{R}$, every subset is compact, so this fails as badly as possible. The correct statement is compact $+$ Hausdorff ambient $\Rightarrow$ closed.

Likewise, two disjoint compact sets in a Hausdorff space have disjoint neighbourhoods — false without it.

What needs a metric

In a metric space: compact $\iff$ sequentially compact $\iff$ complete and totally bounded.

In a general space neither implication holds:

• $[0,1]^{[0,1]}$ with the product topology is compact by Tychonoff and not sequentially compact.
• $[0,\omega_1)$, the first uncountable ordinal with the order topology, is sequentially compact and not compact.

You are unlikely to be asked to prove either, and very likely to be offered the equivalence without the word "metric" in it.

Limit point compactness is a third notion, again equivalent in metric spaces and not in general.

Tychonoff

An arbitrary product of compact spaces is compact in the product topology.

This is equivalent to the axiom of choice. The qualifier matters: it is false for the box topology, which is one more reason the product topology is the default.

Consequences worth having: the Hilbert cube $[0,1]^\omega$ is compact; $\{0,1\}^\omega$ is compact and homeomorphic to the Cantor set; and closed bounded subsets of $\mathbb{R}^n$ are compact, since they sit inside a product of intervals.

Local compactness and compactification

$X$ is locally compact if every point has a compact neighbourhood. $\mathbb{R}^n$ is; $\mathbb{Q}$ is not.

The one-point compactification $X \cup \{\infty\}$ is compact, and is Hausdorff exactly when $X$ is locally compact Hausdorff. Standard instances: $\mathbb{R}\cup\{\infty\}\cong S^1$ and $\mathbb{C}\cup\{\infty\}\cong S^2$, the Riemann sphere from the complex analysis chapters.

Where intuition breaks

"Compact means sequentially compact." Only in metric spaces, and then in both directions. In general neither implies the other.

"Compact subsets are closed." Needs Hausdorff. Cofinite topology is the counterexample and it is extreme: every subset compact, only finite ones closed.

"Closed and bounded means compact." Heine–Borel is $\mathbb{R}^n$. In a general metric space it fails — the closed unit ball of $\ell^2$ — and in a general topological space "bounded" has no meaning at all.

"Tychonoff holds for the box topology." It does not. The theorem is about the product topology, and this is the cleanest reason to care about the distinction.

"An infinite product of compact spaces cannot be compact." Tychonoff says it is. $\{0,1\}^\omega$ being compact is the fact behind the Cantor set.

"Locally compact implies compact." $\mathbb{R}$ is locally compact and not compact. The prefix weakens, as usual.

"The one-point compactification is always Hausdorff." Only when $X$ is locally compact Hausdorff. $\mathbb{Q}\cup\{\infty\}$ is compact and not Hausdorff.

The exam's angle

1. Note the ambient category. Metric, Hausdorff, or arbitrary — each theorem needs one, and the option usually states a weaker one.
2. To disprove compactness, exhibit a cover. In metric spaces a sequence is often easier; in general spaces sequences prove nothing.
3. Cofinite is the first counterexample to try for anything involving compact-implies-closed.
4. For products, Tychonoff answers it — and check the topology is the product one.
5. Compact $+$ Hausdorff is a strong package: normal, and continuous bijections out of it are homeomorphisms. Recognising it saves whole arguments.

The night before

• Compact $=$ every open cover has a finite subcover. Continuous images, closed subsets, finite unions all stay compact.
• Compact $+$ Hausdorff: normal; compact subsets closed; continuous bijection to Hausdorff is a homeomorphism.
• Compact $\iff$ sequentially compact only in metric spaces. $[0,1]^{[0,1]}$ and $[0,\omega_1)$ break it both ways.
• Tychonoff: arbitrary products of compact spaces are compact in the product topology. Equivalent to choice. False for the box.
• Hilbert cube compact; $\{0,1\}^\omega\cong$ Cantor set, compact.
• Locally compact: every point has a compact neighbourhood. $\mathbb{R}^n$ yes, $\mathbb{Q}$ no.
• One-point compactification is Hausdorff $\iff$ $X$ is locally compact Hausdorff. $\mathbb{C}\cup\{\infty\}\cong S^2$.