Prove that the n-cube is connected, for n ≥ 1

[hyderman]

hello

any pone have any idea how to solve this question

thanx

Prove that the n-cube is connected, for n ≥ 1.
(Hint: build the n-cube using two copies of (n − 1)-cube and use induction
on n.)

 

[HallsofIvy]

Think about what they are asking you to do. The "0-cube" is a single point. The "1-cube" is a line segment. You "build the 1-cube using two copies of the 0-cube" by connecting two points. If one of the points is labled "0" and the other is labled "1" then every point on the 1-cube is labled "x" for 0<= x<= 1. I assume that you are allowed to use the fact that the interval [0, 1] is connected.

The 2-cube is a square. You "build the 2-cube using two copies of the 1-cube" by using each one cube as an edge. In particular, if every point on one 1-cube is labled (x, 0) and the other (x, 1), then every point in the 2-cube is labled (x,y) with 0<= y< = 1.

If every point on one n-1-cube is labled (x₁, x₂, ..., x_n-1, 0) and every point on the other n-1-cube is labled (x₁, x₂, ..., x_n-1, 1) then every point on the n-cube is labled (x₁, x₂, ..., x_n-1, x_{n[/sup]). Now use the fact that the n-1-cube and the interval [0,1] is connected to prove that the n-cube is connected.}

 

[Architeuthis Dux]

To prove that the n-cube is connected for n ≥ 1, we can use mathematical induction.

Base Case: For n = 1, the 1-cube is a single point and is trivially connected.

Inductive Hypothesis: Assume that the (n-1)-cube is connected for some n ≥ 1.

Inductive Step: To construct the n-cube, we can start with two copies of the (n-1)-cube and connect them together by adding edges between corresponding vertices. Since we assumed the (n-1)-cube is connected, all vertices in each copy are connected to each other. By connecting the two copies, we ensure that all vertices in the n-cube are also connected. Therefore, the n-cube is connected.

This completes the proof by induction, showing that the n-cube is connected for all n ≥ 1.