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The sequence $n\in [0,\infty )$ is unbounded. Every compact metric space is bounded, by part (b), so this subspace is not compact.

The points $(n,0)$ belong to the indicated subset for every positive integer $n$. They form an unbounded sequence, so this metric space is not bounded. Therefore it is not compact. Shrinking the vertical width does not bound its horizontal extent.

By the Heine-Borel theorem, a noncompact subset $X\subseteq \mathbb{R}$ is either unbounded or not closed. If $X$ is unbounded, the restriction $f(x)=x$ is a continuous function with unbounded range.

Otherwise $X$ is bounded and not closed. Choose $a\in \overline{X}\setminus X$, and defineIts denominator is nonzero on $X$, so it is continuous. Since $a$ is in the closure, for every $n$ there is $x_n\in X$ with $0<|x_n-a|<1/n$. Hence $f(x_n)>n$, proving that this continuous function is unbounded. These two cases establish unbounded continuous witnesses to noncompactness of a real subspace.

We can arrange that neither the supremum nor the infimum is attained. By part (d), choose a continuous function $f$ on $X$ with unbounded range, and put $u=|f|$. Choose $x_n\in X$ so that $t_n=u(x_n)$ increases strictly to infinity. Construct a continuous function $h:[0,\infty )\to \mathbb{R}$ withkeep it constant up to $t_1$ and interpolate linearly on each $[t_n,t_{n+1}]$. Since $t_n\to \infty$, the joining points have no finite accumulation, so this defines a continuous function everywhere on its domain.

Now setBy function composition, $g$ is continuous, and $-1<g(x)<1$ for every $x$. Along the even and odd chosen points,Thus ${\sup }_{X}g=1$ and ${\inf }_{X}g=-1$, and neither is a value of $g$. This proves the bounded continuous witness with neither extremum attained construction, including for subsets that are bounded but not closed.

If $X\subseteq \mathbb{R}$ is not compact, there is an unbounded continuous function $X\to \mathbb{R}$. By the Heine-Borel theorem, either $X$ is unbounded, when $f(x)=x$ works, or it is not closed. In the latter case choose $a\in \overline{X}\setminus X$; the continuous function $f(x)=|x-a{|}^{-1}$ is unbounded because points of $X$ approach the missing limit point $a$. This proof uses the specific compactness characterization of real subspaces; boundedness of a general metric space does not imply compactness.