3

Assume that $K$ is a compact subset of $\mathbb{R}$. Prove directly that $K′=\{(x,0) :x\in K\}$ is a compact subset of $\mathbb{R}^2$.

I have been given the following hint: Suppose that $\{U_i\}_{i\in I}$ is a cover of $K′$ by open sets. These $U_i$ are subsets of the plane $\mathbb{R}^2$. Find a collection $\{V_i\}_{i\in I}$ of open subsets of the real line $\mathbb{R}$ that cover $K$. Be sure to prove that your sets $V_i$ are open—you can’t just say that they are open, you have to prove that.

It really seems kind of like common sense that if all the values of $x$ form a compact set, then $(x,0)$ forms a compact set since $0$ is just $0$. I don't even really understand the hint given.

Housefire
  • 432
  • What would happen if you consider the projection of $U_i$ on $\mathbb{R}$? Will it be open and would cover $K$? – Aniruddha Deshmukh Apr 12 '21 at 05:35
  • Why a projection is an open map? – R. W. Prado Apr 12 '21 at 05:36
  • Basically, the hint is saying to you prove that the projection is an open map. The main idea is that any open $U$ set can be written as a union of open sets of type $(-\delta+x_1,\delta+x_1) \times (-\delta+x_2,\delta+x_2)$ with $(x_1,x_2) \in U$ whose projection into the first coordinate is given by $(-\delta+x_1,\delta+x_1)$, which is open. – R. W. Prado Apr 12 '21 at 05:43
  • The inclusion map is continuous, so apply https://math.stackexchange.com/a/226328/86777 – Joshua P. Swanson Apr 12 '21 at 05:58

2 Answers2

1

Take $\{U_i\}$ open cover of $K’$. Define $V_i=\{x \in R: (x,0) \in U_i\}$. Prove that $V_i$ is open in $R$, i.e., try to find a open ball contained in $V_i$ for all $x\in V_i$.

Hint: exists $r>0$ such that $B((x,0),r)$ is a subset of $U_i$. Use the fact that $|x-y|=||(x,0)-(y,0)||$, where $||.||$ is the euclidean norm in $R^2$

Use compactness of $K$ to obtain a finite subcover, say $V_1, V_2,...,V_k$. Now, take the associated $U_j$.

Uriel Herrera
  • 415
0

Let $\{U_i\}_{i \in I}$ be an open cover of $K'$. Then for each $U_i$ there exists a basic open set $B_i$ such that $B_i \subseteq U_i$. We will show that if $\{B_i\}_{i \in I}$ is an basic open cover of $K'$ then it admits a finite subcover. $B_i$'s are of the form $V_i \times W_i$ where $V_i$, $W_i$ are open sets in $\mathbb{R}$.

For each $x \in K~~~(x,0) \in B_i=V_i \times W_i \Rightarrow x \in V_i$ and $0 \in W_i$ for some $i$. Thus $\{V_i\}_{i \in I}$ is an open cover of $K$, by compactness of $K$ we get $K \subseteq \cup_{i=1}^n V_i$.

Then for any $(x,0) \in K'~~~(x,0) \in \cup_{i=1}^n V_i \times \cup_{i=1}^n W_i=\cup_{i=1}^n (V_i \times W_i)=\cup_{i=1}^n B_i \subseteq \cup_{i=1}^n U_i$. Thus $K' \subseteq \cup_{i=1}^n U_i$.

An easy approach: If one is willing to use the fact that continuous image of a compact set is compact. Then note that the map $f: \mathbb{R} \rightarrow \mathbb{R}^2$ defined by $f(x)=(x,0)$ is continuous and $f(K)=K'$.

absolute0
  • 1,027