---- > [!definition] Definition. ([[complete]]) > It is easy to see that every [[sequence|convergent sequence]] in a [[metric space]] $(X, d)$ is [[Cauchy sequence|Cauchy]].[^1] Should $(X,d)$ satisfy the converse, we say it is **complete**. > Any [[metric space]] can be made complete: see [[completion of a metric space]]. > [!equivalence] > By [[Cauchy sequence is convergent iff has convergent subsequence]], it is equivalent to require that every Cauchy sequence in $(X,d)$ has a [[converge|convergent]] [[subsequence]]. ^equivalence [^1]: Let $x_{1},x_{2},\dots$ be a [[sequence]] in $X$ converging to $L$. Fix $\varepsilon>0$, and pick $N$ large enough that $d(x_{n}, L) < \frac{\varepsilon}{2}$ and $d(x_{m}, L)< \frac{\varepsilon}{2}$ for all $n,m \geq N$. Then $d(x_{n}, x_{m}) \leq d(x_{n}, L) + d(L, x_{m} )< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$, as desired. > [!equivalence] Closed $\iff$ complete for subsets of a metric space > - A [[complete]] [[subspace topology|subspace]] of a [[metric space]] is [[closed set|closed]] > - A [[closed set|closed]] [[subspace topology|subspace]] of a [[complete]] [[metric space]] is [[complete]]. > > > > [!basicexample] Example. > > Early on, one intuits that $\mathbb{Q} \subset \mathbb{R}$ is not complete because 'stuff is missing': there are numbers arbitrarily close to $\mathbb{Q}$ that aren't part of $\mathbb{Q}$. That is precisely the statement that $\mathbb{Q}$ is not closed; this characterization of completeness makes such intuitions precise. > ^basic-example $(1)$ Suppose $Y \subset X$ is a complete subspace of a metric space $X$. Suppose $x_{1},x_{2},\dots$ is a [[sequence]] in $Y$ converging to some $x \in X$. Then $x \in Y$: indeed, $(x_{n})$ is Cauchy in $Y$, and hence converges in $Y$ since $Y$ is complete. In particular, it converges to $x$ since [[limits in Hausdorff spaces are unique]] (and metric spaces are [[Hausdorff space|Hausdorff]]). $(2)$. Suppose $X$ is a [[complete]] [[metric space]] and $Y \subset X$ is a closed subspace. Let $(y_{n})$ be a [[Cauchy sequence]] in $Y$. Since $X$ is complete, $(y_{n})$ converges in $X$, say, to $y \in X$. Since $Y$ is closed, in fact $y \in Y$. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```