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A sequence converges to $x\in X$ whenThe phrase convergent sequence requires the limit to belong to the stated metric space; convergence in a larger ambient space alone does not suffice.

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space.

A subset $U$ of a metric space is open if every $x\in U$ has a radius $r>0$ with the open ball $B_d(x,r)\subseteq U$.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

A subset $B$ of a metric space is bounded if it lies in some ball of finite radius: there are $x_0$ in the space and $R<\infty$ with $d(x,x_0)\le R$ for every $x\in B$. For a nonempty subset, this is equivalent to finite diameter, ${\sup }_{x,y\in B}d(x,y)<\infty$. The empty subset is bounded by convention.

For any topological target $Y$, $f(K)$ is compact. Pull an open cover of $f(K)$ back by $f$, take a finite subcover of $K$, and map the covering statement forward. Without a separation assumption on $Y$, the image need not be closed, as a constant map into a nonclosed singleton of an indiscrete space shows.

If $Y$ is a metric space, $f(K)$ is also closed in $Y$, bounded, totally bounded, complete with its inherited metric, and sequentially compact. Closedness uses the Hausdorff theorem from part (iii); total boundedness follows by taking finite subcovers of the covers by balls of any fixed radius. Compact metric spaces are sequentially compact, and a Cauchy sequence with a convergent subsequence converges to the same limit, proving completeness. These properties are asserted for the image subset, without requiring the whole ambient metric space to be complete.

A metric space is complete if every Cauchy sequence in it converges to a point of that same space. A limit in a larger ambient completion is insufficient for this definition.

A sequence $(x_n)$ in a metric space is a Cauchy sequence ifThis compares sufficiently late terms with each other; it does not assume the existence of a limit in the space.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.

In a metric space,The sequence may equivalently be required to have pairwise distinct terms. The metric hypothesis makes a countable sequence sufficient; arbitrary topological spaces need not have this sequential characterization.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.

If $A$ is closed in the complete ambient metric space, any Cauchy sequence in $A$ converges in $X$ by completeness, and its limit belongs to $A$ by part (i). Thus $A$ is complete with its restricted metric.

Conversely, suppose $A$ is complete and a sequence $a_n\in A$ converges in $X$ to $x$. A convergent sequence is Cauchy, also in the restricted metric, so completeness of $A$ gives $a_n\to a$ for some $a\in A$. Uniqueness of a metric limit gives $x=a\in A$. Part (i) now makes $A$ closed. Hence

A metric space is complete when every Cauchy sequence in it converges to a point of that space. A sequence is Cauchy when, for each $ϵ>0$, there is $N$ such that $d(x_m,x_n)<ϵ$ whenever $m,n\ge N$. The requirement that the limit lie in the space distinguishes completeness from having limits only in a larger ambient space.

If $A$ is closed and $a_n\in A$ converges to $x\in X$, then $x$ cannot lie in the open complement: an open ball about such an $x$ would eventually contain all $a_n$ while missing $A$. Thus $x\in A$.

Conversely, suppose $A$ contains every limit of a sequence in $A$. If $x\in \overline{A}$, every ball $B(x,1/n)$ meets $A$; choose $a_n$ in that intersection. Then $a_n\to x$, so $x\in A$. Hence $\overline{A}=A$, proving that a subset of a metric space is a closed set exactly when it is sequentially closed.

Convergence to a point $x\in X$ meansThe limit must belong to the metric space itself. The definition is equivalent to $d(x_n,x)\to 0$ as a real sequence.

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space. The sequence $x_n=1/(n+1)$ lies in $(0,1)$, but every subsequence converges in $\mathbb{R}$ to zero. Uniqueness of limits excludes an alternative limit inside $(0,1)$, so the open interval is not sequentially compact.

For the product, take any sequence $(x_n,y_n)$. sequential compactness of $X$ gives a subsequence $x_{n_k}\to x\in X$. Apply sequential compactness of $Y$ to the corresponding sequence $y_{n_k}$ to obtain a further subsequence $y_{n_{k_j}}\to y\in Y$. Passing to a subsequence preserves the convergence of the first coordinate, and thereforeThus $X\times Y$ is sequentially compact with the stated sum metric. Taking the second coordinate subsequence inside the first one is essential; two unrelated subsequences would not suffice.

For metric spaces, continuity at $x_0$ means that for every $ϵ>0$ there is $\delta >0$ such thatThe choice of $\delta$ may depend on both $ϵ$ and $x_0$.