Prove sum of line segments with ending points included with euclidean metric is complete iff compact

Question

$A\subset \Bbb{R}$ and let $X(A)\subset \Bbb{R}^2$ be a union of all segments connecting point $(0,1)$ with points $(a,0)$ such that $a\in A$. Show that $X(A)$ with euclidean metric is complete if and only if it is compact.

So ($\Leftarrow$) direction is easy simply notice that for metric spaces compact is equivalent with sequentially compact and if every sequence have convergent subsequence in $X(A)$ then every cauchy sequence is convergent $X(A)$

The other direction ($\Rightarrow$) I wanted to use Bolzano-Weierstrass to prove sequential compactness, but I struggled to prove that $X(A)$ is bounded, because A can be for example uncountable infinite so I think this approach might be wrong. Unfortunately I had no other ideas.

Answer 1

Assume $X(A)$ is complete.

Our goal is to show that $X(A)$ is compact.

If $S$ is a subset of a complete metric space $M$, then $S$ is complete if and only if $S$ is closed.

https://proofwiki.org/wiki/Subspace_of_Complete_Metric_Space_is_Closed_iff_Complete

Hence, since $X(A)$ is complete, and $X(A)$ is a subset of $\mathbb{R}^2$ (which is complete), it follows that $X(A)$ is closed.

Claim:$\;A$ is bounded.

Suppose otherwise.

Without loss of generality, assume $A$ is unbounded to the right.

Let $1 < a_1 < a_2 < a_3,... \in A$ be such that $a_n$ approaches infinity, as $n$ approaches infinity.

Let $P_n=(1,y_n)$ be the point of $\mathbb{R}^2$ where the line segment from the point $(0,1)$ to the point $(a_n,0)$ intersects the vertical line $x=1$.

By definition of $X(A)$, we get that $P_n\in X(A)$, for all $n$.

It's easily shown that the sequence $y_1,y_2,y_3,...$ is a strictly increasing sequence of real numbers such that $y_n$ approaches $1$, as $n$ approaches infinity.

But then, since $X(A)$ is closed, we get $(1,1)$ in $X(A)$, contradiction, since no line segment from the point $(0,1)$ to a point on the $x$-axis passes through the point $(1,1)$.

Hence $A$ is bounded, as claimed.

Then $A\subseteq B$, for some closed, bounded interval $B$.

Since $A\subseteq B$, it follows that $X(A)\subseteq X(B)$.

But $X(B)$ is a closed triangular region, which is clearly bounded.

Hence $X(A)$ is bounded.

Since $X(A)$ is a closed and bounded subset of $\mathbb{R}^2$, $X(A)$ is compact, as was to be shown.

Answer 2

Since $\mathbb{R}^2$ is a complete metric space, the subspace $X(A)$ is complete if and only if it is closed. We shall prove that the following are equivalent:

(1) $X(A)$ is closed.

(2) $X(A)$ is compact.

(3) $A$ is compact.

$(2) \Rightarrow (1)$ is obvious because compact subspaces of Hausdorff spaces are closed.

$(2) \Rightarrow (3)$ is true because $A$ is homeomorphic to the closed subspace $X(A) \cap (\mathbb{R} \times \lbrace 0 \rbrace) = A \times \lbrace 0 \rbrace$ of $X(A)$.

$(3) \Rightarrow (2)$: As you stated, in metric spaces compactness is equivalent to sequential compactness. So let $(x_n)$ be a sequence in $X(A)$. Write $x_n = (s_n,t_n)$. If infinitely many $t_n = 1$, then infinitely many $x_n = (0,1)$ so that we obtain a convergent subsequence. Next consider the case that only finitely many $t_n = 1$, w.l.o.g. all $t_n < 1$. $x_n$ is contained in the segment through $(0,1)$ and $(0,\frac{s_n}{1-t_n})$ where $a_n = \frac{s_n}{1-t_n} \in A$. $(t_n)$ is a sequence in $[0,1]$. It has a convergent subsequence $(t_{n_k})$ with limit $t \in [0,1]$. W.l.o.g. assume that $(t_n) $ converges to $t$. $(a_n)$ is a sequence in $A$. It has a convergent subsequence $(a_{n_k})$ with limit $a \in A$. W.l.o.g. assume that $(a_n) $ converges to $a$. But then $(s_n) = (a_n(1-t_n))$ converges to $a(1-t)$. The point $(a(1-t),t)$ which is the limit of $(x_n)$is contained in the segment through $(0,1)$ and $(a,0)$, i.e. in $X(A)$.

$(1) \Rightarrow (3)$: Assume $A$ is not compact. If it is not closed in $\mathbb{R}$, then $X(A)$ cannot be closed in $\mathbb{R}^2$ (consider $A \times \lbrace 0 \rbrace$ which is closed in $X(A)$). Therefore it suffices to consider a closed $A$. We conclude that $A$ is not bounded. Let $(a_n)$ be a sequence in $A$ such that $\lvert a_n \rvert \to \infty$; w.l.o.g. $a_n \to \infty$ and all $a_n >1$. Let $x_n = (s_n,t_n)$ be a point on the segment through $(0,1)$ and $(a,0)$ such that $s_n = 1$. Then necessarily $t_n \to 1$. But $(1,1)$ which is the limit of $(x_n)$ is not contained in $X(A)$.

The same is true for any subset $A \subset \mathbb{R}^n$. In this case $X(A) \subset \mathbb{R}^{n+1}$. It is easy to see that $X(A) - \lbrace (0,1) \rbrace$ is homeomorphic to $A \times [0,1)$. If $A$ is compact, $X(A)$ is the one-point-compactification of $A \times [0,1)$ and can be identified with $A \times [0,1]/A \times \lbrace 1 \rbrace$. The latter is known as the cone $CA$ of $A$. Note that we always have a continuous bijection $b : X(A) \to CA$; it is a homeomorphism if and only if $A$ is compact.

Answer 3

Here is an alternative proof using some basic facts about compact spaces (general topology) and normed linear spaces. It illuminates the difference between the finite-dimensional and the infinite-dimensional situation.

For any normed linear space $E$ over the real field and any $A \subset E$ define $X(A) \subset E \times \mathbb{R}$ as the union of all line segments connecting the point $(0,1)$ with a point $(a,0)$ such that $a \in A$. Note that $ E \times \mathbb{R}$ is again a normed linear space (take e.g. $\lVert (x,t) \rVert = \lVert x \rVert + \lvert t \rvert$).

Let $E' = E \times [0,1]$. Define $r : E' \to E', r(x,t) = ((1-t)x,t)$. This is a continuous function and $r(\lbrace x \rbrace \times [0,1])$ is the line segment connecting $(0,1)$ with $(x,0)$, hence $r(A \times [0,1]) = X(A)$. Obviously $r$ restricts to a homeomorphism $\rho : E \times [0,1) \to E \times [0,1)$.

Let $E_0 = E \times \lbrace 0 \rbrace$. Define $h : E \to E_0, h(x) = (x,0)$. This is a homeomorphism. Observe that $A_0 = A \times \lbrace 0 \rbrace = X(A) \cap E_0$ is a closed subspace of $X(A)$, and if $X(A)$ is closed in $E \times \mathbb{R}$, then $A_0$ is closed in $E_0$ (clearly the latter holds if and only if $A$ is closed in $E$).

Let us first show that the following are equivalent:

(1) $A$ is compact.

(2) $X(A)$ is compact.

$(1) \Rightarrow (2)$ $A \times [0,1]$ is compact, hence $r(A \times [0,1] = X(A)$ is compact.

$(2) \Rightarrow (1)$ $A_0$ is closed in $X(A)$, hence it is compact. Therefore $A = h^{-1}(A_0)$ is compact.

In finite-dimensional normed linear spaces it is known that the compact subsets agree with the closed bounded subsets. This is definitely wrong in infinite-dimensional normed linear spaces (although compact subsets are always closed and bounded).

Let us next show that the following are equivalent:

(1) $A$ is closed and bounded.

(2) $X(A)$ is closed and bounded.

It is obvious that $A$ is bounded if and and only if $X(A)$ is bounded.

$(1) \Rightarrow (2)$ Since $X(A) \subset E'$ and $E'$ is closed in $E \times \mathbb{R}$, it suffices to show that $X(A)$ is closed in $E'$. We show that $W = E' \backslash X(A)$ is open. Obviously $V = \rho((E \backslash A) \times [0,1))$ is open in $E \times [0,1)$, hence also open in $E'$. By construction it contains all $(x,t) \in W$ such that $t < 1$. Now consider a point $(x,1) \in W$. We must have $x \ne 0$. Since $A$ is bounded, there exists $M > 0$ such that $\lVert a \rVert \le M$ for all $a \in A$. Assume that each open neighborhood of $(x,1)$ intersects $X(A)$. In particular we find $(\xi,\tau) \in X(A)$ such that $\lVert \xi \rVert > \frac{\lVert x \rVert}{2}$ and $\tau > 1 - \frac{\lVert x \rVert}{2M}$. $(\xi,\tau)$ is contained in the line segment connecting $(0,1)$ and $(\frac{\xi}{1-\tau},0)$ with $\frac{\xi}{1-\tau} \in A$. But $\lVert \frac{\xi}{1-\tau} \rVert > \frac{\lVert x \rVert}{2}/\frac{\lVert x \rVert}{2M} = M$ which is a contradiction. Hence $(x,1)$ has an open neighborhood contained in $W$.

$(2) \Rightarrow (1)$ We know that $X(A)$ closed implies $A$ closed.

If $E$ is finite-dimensional (i.e. essentially $E = \mathbb{R}^n$), we finally show that the following are equivalent:

(1) $A$ is compact.

(2) $X(A)$ is closed.

$(1) \Rightarrow (2)$ follows from our first result.

$(2) \Rightarrow (1)$ Let $S^E = \lbrace x \in E \mid \lVert x \rVert = 1 \rbrace$; this is a compact set. $S(A) = X(A) \cap S^E \times [0,1]$ is a closed subset of the compact $S^E \times [0,1]$, hence it is itself compact. Since $X(A)$ does not contain any point $(x,1)$ except for $x = 0$, we have $S(A) \subset S^E \times [0,1)$ so that $S(A) \subset S^E \times [0,\tau]$ for some $\tau \in [0,1)$. Let $a \in A$. The point $(\frac{a}{\lVert a \rVert},1 - \frac{1}{\lVert a \rVert})$ lies in $S(A)$. Therefore $1 - \frac{1}{\lVert a \rVert} \le \tau$ which means $\lVert a \rVert \le \frac{1}{1-\tau}$. Hence $A$ is bounded. That $A$ is closed is known.

If $E$ is infinite-dimensional, then there exist non-compact such that $X(A)$ is closed. Of course $A$ must be closed. E.g. $S^E$ is closed and non-compact (although bounded) and $X(A)$ is closed. There are also examples of closed unbounded $A$ such that $X(A)$ is closed.