Completeness and Baire category

In one sentence

Completeness is a property of the metric, not of the topology, and both of the theorems built on it — Banach's fixed point theorem and Baire's category theorem — fail loudly when their hypotheses are trimmed.

Why the exam asks it

Because "complete" sounds like an intrinsic property of a space, and it is not. $(0,1)$ and $\mathbb{R}$ are homeomorphic; one is complete and the other is not. That single fact turns a whole family of natural-sounding options false, and it is examined because it genuinely surprises people.

The idea

A sequence is Cauchy if for every $\varepsilon > 0$ there is an $N$ beyond which all terms are within $\varepsilon$ of each other. $X$ is complete if every Cauchy sequence in $X$ converges in $X$.

Convergent always implies Cauchy. The reverse is exactly what completeness asserts, and the point of the definition is that Cauchyness can be checked without knowing the limit.

$\mathbb{R}$ and $\mathbb{R}^n$ are complete. $\mathbb{Q}$ is not: a sequence of rationals converging to $\sqrt2$ is Cauchy in $\mathbb{Q}$ with nowhere to land. $(0,1)$ is not, since $1/n$ is Cauchy and its limit is missing.

Completeness is not topological

$(0,1)$ is homeomorphic to $\mathbb{R}$ — for instance by $x \mapsto \tan(\pi(x - 1/2))$. One is incomplete, the other complete. So completeness cannot be determined from the open sets alone; it depends on the metric, and a homeomorphism can move the metric.

Compactness, connectedness and separability are topological. Completeness, boundedness, total boundedness and uniform continuity are not. Sorting the properties into those two lists is worth more marks than any single theorem in this topic.

Two standard relationships

• A closed subset of a complete space is complete.
• A complete subset of any metric space is closed.

These are converses of each other in the right setting, and both directions are used.

Cantor's intersection theorem

In a complete metric space, if $F_1 \supseteq F_2 \supseteq \cdots$ are non-empty closed sets with $\operatorname{diam}(F_n) \to 0$, then $\bigcap F_n$ is a single point.

Drop the diameter condition and the conclusion fails outright: $F_n = [n, \infty)$ in $\mathbb{R}$ are non-empty, closed and nested, and no real number lies in all of them. Drop completeness and it fails too. Both hypotheses are load-bearing, and both get removed in options.

Banach's fixed point theorem

$X$ complete, $f : X \to X$ with $d(f(x), f(y)) \le k\, d(x,y)$ for a fixed $k < 1$. Then $f$ has exactly one fixed point, and iterating from any starting point converges to it.

The hypothesis is a uniform constant $k < 1$. The weaker-looking condition $d(f(x),f(y)) < d(x,y)$ for all $x \ne y$ is genuinely weaker and is not enough:

$$f(x) = x + \frac1x \quad \text{on } [1,\infty).$$

Here $f'(x) = 1 - 1/x^2$, which lies in $[0,1)$ throughout, so by the mean value theorem $|f(x)-f(y)| < |x-y|$ whenever $x \ne y$. The domain is closed in $\mathbb{R}$, hence complete. And $f$ has no fixed point at all: $f(x) = x$ forces $1/x = 0$, which no real number satisfies.

The reason is visible in the derivative — it approaches $1$ without ever reaching it, so no single $k < 1$ dominates it. That gap between "less than at each pair" and "uniformly less than" is the whole example.

Baire's category theorem

Call $A$ nowhere dense if $\overline{A}$ has empty interior, and meagre (first category) if it is a countable union of nowhere dense sets.

A complete metric space is not meagre in itself. Equivalently: a countable intersection of dense open sets is dense.

Consequences worth carrying:

• $\mathbb{R}$ is uncountable. If it were countable it would be a countable union of singletons, each nowhere dense.
• $\mathbb{Q}$ is not a $G_\delta$ in $\mathbb{R}$ — not a countable intersection of open sets. Suppose it were; $\mathbb{R}\setminus\mathbb{Q}$ would be a countable union of closed nowhere dense sets, and adding the singletons $\{q\}$ would make $\mathbb{R}$ meagre in itself.
• Therefore no function is continuous exactly on $\mathbb{Q}$, because the set of continuity points of any function is always a $G_\delta$.
• The reverse is possible: Thomae's function is continuous exactly on the irrationals, and the irrationals are a $G_\delta$. The asymmetry is real and is a favourite question.

$\mathbb{Q}$ itself is meagre in itself, which is not a contradiction — it is simply not complete.

Where intuition breaks

"$(0,1)$ and $\mathbb{R}$ are homeomorphic, so both are complete or neither is." Completeness is not a topological invariant. The most productive single fact in this topic.

"Every Cauchy sequence converges." Only in a complete space. In $\mathbb{Q}$ the truncated decimals of $\sqrt2$ are Cauchy and have no limit.

"$d(f(x),f(y)) < d(x,y)$ for all $x \ne y$, so $f$ has a fixed point." $x + 1/x$ on $[1,\infty)$. Tempting because the condition looks like a contraction and, on a compact domain, it would suffice. Completeness is not compactness.

"Nested non-empty closed sets have non-empty intersection." $[n,\infty)$. The diameter condition is the hypothesis, and it is the one options delete.

"A countable union of nowhere dense sets can be all of $\mathbb{R}$." Baire says no. This is the form in which category theory is usually tested — not by name, but as a claim about writing a complete space as a countable union of small sets.

"There is a function continuous exactly at the rationals." No. Continuity sets are $G_\delta$, and $\mathbb{Q}$ is not. Continuous exactly at the irrationals is fine — Thomae. The pair is set as a two-option question and the answer is not symmetric.

"A complete and bounded space is compact." Bounded is not totally bounded. The closed unit ball of $\ell^2$ is complete and bounded and not compact.

The exam's angle

1. Sort the property: topological or metric? Compact, connected, separable: topological. Complete, bounded, totally bounded, uniformly continuous: metric. Any option pushing a metric property through a homeomorphism is false.
2. For fixed-point options, look for the uniform constant. If the option says only "strictly decreases distances", reach for $x + 1/x$ on $[1,\infty)$.
3. For nested sets, check the diameters. If they do not shrink, expect the intersection to be empty and look for $[n,\infty)$.
4. For anything phrased as a countable union covering a space, think Baire. Then decide whether the space is complete, because that is the hypothesis.
5. Continuity sets are $G_\delta$. That single fact answers every "continuous exactly on…" question the paper can pose.

The night before

• Complete: every Cauchy sequence converges. $\mathbb{R}$, $\mathbb{R}^n$ yes; $\mathbb{Q}$, $(0,1)$ no.
• Completeness is not topological. $(0,1) \cong \mathbb{R}$, only one is complete.
• Closed in complete $\Rightarrow$ complete. Complete $\Rightarrow$ closed.
• Cantor: nested, closed, non-empty, diam $\to$ 0, complete $\Rightarrow$ one common point. $[n,\infty)$ shows the diameter hypothesis is essential.
• Banach needs a uniform $k < 1$. $x + 1/x$ on $[1,\infty)$: distances strictly decrease, no fixed point.
• Baire: a complete metric space is not a countable union of nowhere dense sets.
• $\mathbb{Q}$ is not $G_\delta$; no function is continuous exactly on $\mathbb{Q}$. Thomae is continuous exactly on the irrationals.
• Complete + totally bounded $=$ compact. Complete + bounded does not.