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

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The n-cube is proven to be connected for n ≥ 1 by utilizing induction and the properties of lower-dimensional cubes. The construction of the n-cube involves two copies of the (n-1)-cube, where each point on the (n-1)-cube is labeled to extend into the n-dimensional space. Specifically, the 0-cube is a point, the 1-cube is a line segment, and the 2-cube is a square, with each higher dimension built upon the previous one. The connectedness of the n-cube follows from the connectedness of the (n-1)-cube and the interval [0, 1].

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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.)
 
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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, ..., xn-1, 0) and every point on the other n-1-cube is labled (x1, x2, ..., xn-1, 1) then every point on the n-cube is labled (x1, x2, ..., xn-1, xn[/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.
 

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