No Holes Allowed! What's a Complete Space?

by Marcus Math

🔍

### Prerequisites: 
- [Closed, Open, Clopen](https://www.noteblogdoku.com/blog/closed-open-clopen)
- [Metric Space vs Inner Product Space](https://www.noteblogdoku.com/blog/metric-space-vs-inner-product-space)

## What is a Complete Space?

A Complete Space is defined such that for a given metric space, there are no holes. This is fundamental to the definition of a Hilbert Space (complete inner product space) which behave like infinite dimensional Euclidean Spaces. Hilbert spaces are fundamental to Quantum Mechanics, Fourier Analysis, Signal Processing, and Machine Learning.

One may ask why the condition that it must be *complete* is important. Although an inner product gives you geometry (lengths, angles, orthogonality, & projections) most everything you want to use that geometry for involves taking limits (perhaps implicitly). Completeness guarantees those limits exist within the space.

To show that a Metric Space is Complete, we are going to introduce 
- Convergent Sequences
- Cauchy Sequences

and show that when all Cauchy Sequences are Convergent a Metric Space that this Metric Space is Complete.

What? That's a lot of jargon, anyway let's begin.

## Convergent Sequences

A sequence $(x_n)$ in a metric space $(X,d)$ **converges** to a point $x \in X$ if for every $\epsilon > 0$, there exists a natural number $N$ such that $\forall n \geq N$

$$ d(x_n,x) < \epsilon$$

This is stating that no matter how small a positive distance $\epsilon$ you choose, eventually (some number greater than $N$) every term of the sequence lies within some distance $\epsilon$ of the limit point $x$.

Another way to restate this is:
$$ \forall \epsilon > 0\  \exists N \in \mathbb{N}\ \forall n \geq N: d(x_n,x) < \epsilon $$

The point $x$ is called the **limit** of the sequence writen $x_n \rightarrow x$ or the canonical form (thanks calculus!)

$$ lim_{n \rightarrow \infty} x_n = x$$
Sometimes $x$ is denoted as Limit $L$, giving us

$$ lim_{n \rightarrow \infty} x_n = L$$

#### Unique Limits Quick Aside:
If $x_n \rightarrow x$ and $x_n \rightarrow y$ then $\forall \epsilon$, $d(x,y) \leq d(x,x_n) + d(x_n,y) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ so $d(x,y) = 0$. Thus $x=y$ by the positive definiteness of the metric. Note that $\frac{\epsilon}{2}$ statement in the triangle inequality comes from our definition of convergent sequence where we just make $\epsilon \rightarrow \frac{\epsilon}{2}$ which is logically valid.

For some additional useful terminology for sequences in metric space $(X,d)$
- **Convergent Sequences** converge to a limit $L \in X$
- **Divergent Sequences** do not converge

Another way to state this definition is that a convergent sequence $x_n$ approaches the limit $L$.

**NOTE** the limit $L \in X$ must exist! (the limit $L$ must be present in the set $X$ that defines our metric space $(X,d)$. From first principles, we are measuring the distance between an element in the sequence $x_n$ and $L$, and that this distance is less than $\epsilon$:
$$ d(x_n,L) < \epsilon$$

For this distance function to be valid the condition $L \in X$ for the metric space $(X,d)$ is necessary.

<img src="/media/images/20260604_074814_06_04_2026_00_47_01_convergent_sequence_animation.gif" alt="06 04 2026 00 47 01 convergent sequence animation" style="max-width: min(800px,85%); height: auto;">
<center>
$\textbf{Figure 1}$: Example of a Convergent Sequence that Approaches $L=1$

Shows that for any given $\epsilon > 0$ only finitely many terms of the sequence escape $(L- \epsilon, L+ \epsilon)$
</center>

## Cauchy Sequences

A sequence $(x_n)$ in a metric space $(X,d)$ is **Cauchy Sequence** if for every $\epsilon > 0$ $\exists N$ such that
$$ d(x_m,x_n) < \epsilon \quad \forall m,n \geq N$$

**Note** that this definition doesn't mention limits, and *only* mentions the distance between all pairs of items in the sequence after term $x_n$ (assuming $m>n$). Intuitively, a sequence is Cauchy if its terms bunch together as the sequence progresses. Past some finite index $N$, *every* pair of terms (not just consecutive ones) lies within distance $\epsilon$ of each other. Thus, the entire tail of the sequence gets arbitrarily close together.

This is exactly what the definition states: 
By choosing $\epsilon > 0$ as small as we like, we can always find a point $N$ beyond which 
$$d(x_m, x_n) < \epsilon \quad \forall m, n \geq N$$

This also leads into a useful theorem stating that:
*Cauchy sequences are bounded*

To prove this, take $\epsilon =1$, to get an $N$ with $d(x_m,x_n) < 1$, $\forall m,n \geq N$. By definition of Cauchy, every term greater than $N$ lies within distance $1$ of the single anchor point $x_N$, except the finitely many preceding terms $x_1,... x_{N-1}$.

For these finitely many terms $x_1,...x_{N-1}$ set the maximum distance $M$ to be the largest distance between these $N-1$ terms (and 1) and anchor point $x_N$
$$ M = \max\{1, d(x_1,x_N),d(x_2,x_N),...d(x_{N-1},x_N)\} $$

This is a maximum over finitely $N$ real numbers, with every distance term satisfying
$$ d(x_k, x_N) \leq M$$
Thus the entire sequence lies in the closed ball $B(x_N,M)$ and is bounded.

<img src="/media/images/20260604_083842_06_04_2026_01_37_28_cauchy_sequence_animation.gif" alt="06 04 2026 01 37 28 cauchy sequence animation" style="max-width: 100%; height: auto;">
<center>
$\textbf{Figure 2}$: The Sequence of Decimal Truncations of $\sqrt{2}$ in $\mathbb{Q}$ (the rationals): a Cauchy sequence with no limit in $\mathbb{Q}$.
</center>

#### Example of Cauchy Sequence: Decimal Truncations of $\sqrt{2} \in \mathbb{Q}$

Given the rationals $\mathbb{Q}$ 
$$ \mathbb{Q} = \left\{ \frac{p}{q} \;:\; p, q \in \mathbb{Z},\ q \neq 0 \right\} $$

We are going to show that the sequence shown in **Figure 2**, the Decimal Truncations of $\sqrt{2}$ is Cauchy in $\mathbb{Q}$. As we know $\sqrt{2} \notin \mathbb{Q}$ ($\sqrt{2}$ irrational), however $\sqrt{2} \in \mathbb{R}$. The key result that we will show is that *a Sequence can be Cauchy even if its limit is not in the set*. Recall that the definition of Cauchy does not include the limit!

#### The Sequence

Define $x_n$ to be the truncation of $\sqrt{2}$ to $n$ decimal places:
$$ x_n = \frac{\lfloor 10^n \sqrt{2} \rfloor}{10^n}, \qquad n = 0, 1, 2, \dots $$
giving $x_0 = 1,\ x_1 = 1.4,\ x_2 = 1.41,\ x_3 = 1.414,\ \dots$ as in **Figure 2**.
By construction, each $x_n$ approximates $\sqrt{2}$ from below to within one unit in the
last place:
$$ \text{Eq. 1: } \quad 0 \leq \sqrt{2} - x_n < 10^{-n} $$

**Every term is rational.** 
Each $x_n$ is an integer divided by $10^n$, so $x_n \in \mathbb{Q}$. The entire sequence therefore lives inside $\mathbb{Q}$.

**The sequence is Cauchy in $\mathbb{Q}$.** 
Let $\epsilon > 0$. By the Archimedean property (see **Appendix**) we may choose $N$ with $10^{-N} < \epsilon$. Take any $m, n \geq N$. By Eq. 1, both $x_m$ and $x_n$ lie in the interval
$$\left(\sqrt{2} - 10^{-N},\ \sqrt{2}\,\right]$$

which has length $10^{-N}$. Two points in an interval of that length differ by less than its length, so
$$ d(x_m, x_n) = |x_m - x_n| < 10^{-N} < \epsilon. $$
Since $\epsilon > 0$ was arbitrary, $(x_n)$ satisfies the Cauchy condition.

**The sequence has no limit in $\mathbb{Q}$.** 
From our previous steps we have
$$|\sqrt{2} - x_n| < 10^{-n} \to 0$$

so $x_n \to \sqrt{2} \in \mathbb{R}$. Suppose, for contradiction, that $x_n \to q$ for some $q \in \mathbb{Q}$. Then $(x_n)$ would
converge to $q$ in $\mathbb{R}$ as well. However, limits in $\mathbb{R}$ are unique, forcing $q = \sqrt{2}$. This contradicts $\sqrt{2} \notin \mathbb{Q}$. Thus no rational limit exists.

#### Conclusion

The sequence $(x_n)$ is **Cauchy in $\mathbb{Q}$** yet has **no limit in $\mathbb{Q}$**. This proves a key claim: *a sequence can be Cauchy without converging in the space*, because the Cauchy condition only constrains distances between terms. It never refers to a limit point.

Equivalently, this shows **$\mathbb{Q}$ is not complete**. The same sequence does converge in $\mathbb{R}$ (to $\sqrt{2}$), showing that completeness is a property of the *space*, not of the sequence.

## How are Cauchy & Convergent Sequences Related?

To recap Cauchy and Convergence Sequences
- **Convergence** measures distances between the terms $(x_n)$ and some limit $L$
- **Cauchy** measures distances between the terms themselves

Naturally, one may notice that they are similar since both seem to converge to a point. Here we are going to prove that 
$$ \text{Every Convergent Sequence is a Cauchy Sequence} $$

and that this is true in any metric space $(X,d)$. Note that that the converse of this statement is not true as seen by **Figure 2**.

**Theorem.** Let $(X, d)$ be a metric space. If $(x_n)$ converges to some $x \in X$, then $(x_n)$ is Cauchy.

**Proof.** Let $\epsilon > 0$. Since $x_n \to L$, the definition of convergence lets us apply it with the halved tolerance $\tfrac{\epsilon}{2} > 0$: there exists an $N$ such that
$$ d(x_n, L) < \frac{\epsilon}{2} \qquad \forall n \geq N $$

Now take any two indices $m, n \geq N$. Both terms are within $\tfrac{\epsilon}{2}$ of the limit $L$, so by the triangle inequality (which we get from properties of metric space $(X,d)$)
$$ d(x_m, x_n) \;\leq\; d(x_m, L) + d(L, x_n) \;<\; \frac{\epsilon}{2} + \frac{\epsilon}{2}
\;=\; \epsilon. $$
Since $\epsilon > 0$ was arbitrary, $(x_n)$ satisfies the Cauchy condition.

To restate, the Cauchy condition is that $\forall \epsilon >0$, we can always find a point $N$ beyond which $d(x_m,x_n) < \epsilon$, $\forall m,n \geq N$. This is exactly what we showed.

Note that for the $\frac{\epsilon}{2}$ trick that we can feed any positive number into the definition of convergence. Choosing $\frac{\epsilon}{2}$ makes the two halves add back to exactly $\epsilon$. If $\epsilon$ was used instead of $\frac{\epsilon}{2}$, then we'd get $d(x_m,x_n) < 2 \epsilon$, which logically still works, but we'd have to rename $2\epsilon$ to $\epsilon$.

With this proof, convergence gives us an anchor point limit $L$ that every later term is close to. Closeness to this common point $L$ forces the terms of the sequence to be close to each other. This is basically what the triangle inequality is stating.

We saw in **Figure 2** that the reverse is not true, since although the sequence becomes arbitrarily close to some point $L$, that $L \notin X$ thus $d(x_i,L)$ is invalid $\forall x_i \in X$ since the distance function for a metric space is defined only on the elements $x_i \in X$.

## What's a Complete Space?

As teased in the beginning a **Complete Space** is a space without holes, but what does this mean?

$$ \text{A metric space is complete when every Cauchy sequence in it converges to a Limit that is also in the space}$$

Another way to state this is that

$$ \textbf{A metric space is complete when every Cauchy sequence in it is also a Convergent Sequence} $$

Some examples of complete spaces include $\mathbb{R}$, $\mathbb{R}^n$, $\mathbb{C}$, and $C[a,b]$ with the sup norm., and some examples of non-complete spaces are $\mathbb{Q}$, $(0,1)$, and $\mathbb{R} \setminus \{0\}$. The key thing in common between the non-complete spaces is that there are points not in the space that exist such that a sequence inside the space can approach it but never reach. We saw an example for $\mathbb{Q}$ in **Figure 2**, but consider $(0,1)$. For $(0,1)$, we could have a sequence that approaches 0, however since $0 \notin (0,1)$, the limit of that sequence is not in the space. Likewise for $\mathbb{R} \setminus \{0\}$, this is the reals with a hole at 0. In this space too, if there is a sequence that has a limit of 0, since $0 \notin \mathbb{R} \setminus \{0\}$, we get that this space is not complete.

### Why do Complete Spaces not have holes?

From the definition of a Cauchy sequence, we can see that the definition is defined in terms of the terms of the sequence growing closer together over time, and does not mention any limit points. So a Cauchy sequence is a way to narrow down the position of a sequence using only the terms of the sequence, since as we go further out, the terms of the sequence are constrained to smaller and smaller balls. In effect, it behaves like a sequence that converges, however there is no requirement that this has to.

The two possibilities for a Cauchy sequence are
1. There is a point (limit) in $X$ at the location it's zooming in on. Then the sequence converges to that point 
2. There is not a point in $X$ at the location that the sequence is converging to
    - There exists a **hole** that exists, which is an empty location that sequences converge upon

A space is complete precisely when possibility 2 never happens — that is, when every Cauchy sequence falls into case 1. Equivalently:
$$ \text{X is complete} \leftrightarrow \text{every Cauchy sequence in X converges in X} \leftrightarrow \text{X has no holes}$$

This is why the two statements of the definition say the same thing. "Every Cauchy sequence converges" is just the precise way of saying "every location the space's own sequences zoom in on is actually filled," and a missing point at such a location is a hole.

---
## Appendix

### Archimedean Property
Formal Definition
Let $(F, +, \cdot, <)$ be an ordered field (such as the rational numbers $  \mathbb{Q}  $ or the real numbers $  \mathbb{R}  $).
The field is called Archimedean if it satisfies the following property:
For any two positive elements $  x, y \in F  $ with $  x > 0  $ and $  y > 0  $, there exists a natural number $  n \in \mathbb{N}  $ such that $ n \cdot x > y$

I.e. no matter how small $  x  $ is or how large $  y  $ is (as long as both are positive), you can always multiply $  x  $ by a large enough integer to make it bigger than $  y  $.

### Ordered Field 
An ordered field is a mathematical structure that combines two things:

1. The algebraic structure of a field
    - Add, multiply, subtract, and divide
2. A total order $<$ that is compatible with field operations
    - Ordering means you can compare any two elements
    - For any two numbers, one is larger, smaller, or equal

Examples include the reals $\mathbb{R}$ and the rationals $\mathbb{Q}$. 
**NOTE** that not all Fields have to be ordered, such as the complex numbers $\mathbb{C}$ since this field cannot be totally ordered while preserving addition/multiplication.