Showing that a set is closed, bounded, and non-compact

[A Tripolation]

To begin with, I'm in my first real mathematics course this semester. Most of my work has been concentrated in physics and I have no background in analysis and only some basic set theory.

 

The question is:

 

Let [latex] R^\infty[/latex] be the set of all infinite-tuples [latex](x_1, x_2,...)[/latex] of all real numbers that end in an infinite string of 0's. Define an inner product via

 

[latex] \left\langle x,y \right\rangle = \sum_{i=1}{^\infty } x_i y_i[/latex]

 

Let ||x-y|| be the induced metric on [latex] R^\infty[/latex]. Let [latex] e_i[/latex] be the vector with a 1 in the ith entry and a 0 in the other entries. These form a basis for [latex] R^\infty[/latex].

 

Let X be the set {[latex]e_i[/latex]}[latex]_{1\leq i \leq \infty}[/latex]. Show that X is closed, bounded, and non-compact.

 

I understand the formulation of the summation for the dot product and the formation of the basis using those vectors (it forms a linearly independent set that spans the space, IIRC my algebra correctly), but I don't know where to start showing that this set is closed. Or bounded (why would an infinite set be bounded?). I have some notes on compactness, but the concept is quite foreign to me. Any further explanations on the topic would be quite beneficial to me.

 

Thanks!

[ajb]

By definition a closed set is one whose complement is an open set. You have a metric and so a topological space, so this should make sense.

 

For boundedness, you have a metric space and so can consider balls. You need to think of subsets contained in finite radius balls.

 

A metric space is compact if it is complete and (totally) bounded.

 

I hope that gives you some clues.