0
$\begingroup$

Let us define $I(a,b)$ as a line segment between $a,b\in \mathbb{R}^n$ ($(a,b) \in I(a,b)$)

There is a family of problems where there are two sets in $\mathbb{R}^n$ and both of them have some topological property and the task is to show that the family of segments between those two sets have the same properties.

For example

(1) We have two sets $A \subset \mathbb{R}^+$, $B \subset \mathbb{R}^+$ and $A, B$ are both compact (or connected) and we have to show that $I(A,B) = \bigcup\{I((a,0),(0,b)): a\in A, b\in B\} \subset \mathbb{R^2}$ is compact (connected)

(2) $A \subset \mathbb{R}^2\times\{0\}$, $f: A \rightarrow \mathbb{R}$ and $f$ continous let $$X = \bigcup \{I(a,b): b=(0,0,f(a)), a \in A\}$$ prove that if A is compact (or connected) then X is compact (or connected).

How can I prove statements like these? Which tools (or theorems) are useful in these kind of proofs?

2 Answers 2

1

Images and products of compact sets are compact, and images and products of connected sets are connected. Then, because $[0,1]$ is compact and connected, it follows:

1) $I(A,B)$ is exactly the image of the continuous function $A\times B\times [0,1] \to \mathbb{R}^2$ sending $(a,b,\lambda)$ to $\lambda (a,0) + (1-\lambda)(0,b)$. Therefore, if $A$ and $B$ are compact, $I(A,B)$ is compact.

2) $X$ is the image of the continuous function $A\times B\times [0,1]\to \mathbb{R}^3$ sending $((a_1,a_2),b,\lambda)$ to $\lambda (a_1,a_2,0) + (1-\lambda)(0,0,f(a))$. Therefore, if $A$ and $B$ are connected, $X$ is connected.

1

I'd use brute force. The cases you've written are easy to imagine (and even draw) so for compactness you should prove closure and boundness, and for connectedness, path connectedness (build the function) e.g.:

(1) Compacness: As $A,B$ are bounded, we can define $M=\sup(A\cup B),m=\inf(A\cup B)$ and as they're close $M,m$ are reached, so $I(A,B)=[m,M]$ which is compact.

Connectedness: Set $x,y\in I(A,B)$. As $I(x,y)\subset I(A,B)$ we can define $f:[0,1]\rightarrow I(A,B),f(\lambda)=\lambda y+(1-\lambda)x$ which is continuous.

  • 0
    I don't get it. $I(A,B)$ is obviously bounded but why is it closed subset? – 2017-02-16
  • 0
    As I said since $A\cup B$ is closed $m,M\in A\cup B\subset I(A,B)$, so by definition if $m\in A,M\in B$, or $m\in B,M\in A$ then $[m,M]=I(m,M)\subset I(A,B)$. Otherwise, without loss of generality if $m,M\in A$, then define $I_1=I(m,\sup B),I_2=I(\inf B,M)$ and by hypothesis $I_1,I_2$ must overlap so as $I_1\cup I_2=[m,M]$ and both by definition are subsets of $I(A,B)$, then $[m,M]\subset I(A,B)$ – 2017-02-16
  • 0
    O! I probably had written something different than I thought The problem is a bit different: Let's put A on X-axis and B on Y-axis and I(a,b) is line segment in $R^2$ between a on X-axis and b on Y-axis (in other words I meant $I((a,0),(0,b))$ – 2017-02-16