# Thread: Every infinite bounded set of R^n has an accumulation point

1. ## Every infinite bounded set of R^n has an accumulation point

So I have a general idea of what the proof should like (construct a sequence in S, since it is bounded the sequence has a convergent subsequence, and so the limit of that subsequence is an accumulation point). However, I am having a hard time construction such a sequence such that none of the terms of the subsequence are the limit point. How would I go about doing this. I considered using the supremum of S as the limit point but if the supremum is in S, there is no guarantee that every single term of a constructed sequence that converges to the supremum has terms in which none of the terms are equal to the supremum itself. Any help would be appreciated, thanks.

2. Okay, so proving this in $\mathbb{R}$ is pretty easy but carrying that over to $\mathbb{R}^{n}$ is another thing altogether. I am tempted to just apply the infinite and bounded condition to every component of each term in a sequence in $\mathbb{R}^{n}$, but just because the set is infinite in $\mathbb{R}^{n}$ doesn't mean that if we look at each component and the set of all possible elements for that component is infinite, for instance the set (x,y,z) where x = y = 1 and z in [-1,1] is bounded and infinite but there are not infinitely many choices for each component. How do I resolve this?

3. Since $S$ is bounded, then there is a rectangle $R_1=[a_{11},b_{11}]\times[a_{12},b_{12}]\times\cdots\times[a_{1n},b_{1n}]$ which includes $S$. Partition this rectangle into $2^n$ equally-sized rectangles. In particular,

$P_i=\{I_1\times I_2\times\cdots\times I_n:I_j\in\{[a_{ij},(a_{ij}-b_{ij})/2],[(a_{ij}-b_{ij})/2,b_{ij}]\}\}$,

where $i=1$.

Since there are finitely many ( $2^n$) rectangles in $P_i$, at least one must contain infinitely many points of $S$. Choose that one, and denote it $R_2=[a_{21},b_{21}]\times[a_{22},b_{22}]\times\cdots\times[a_{2n},b_{2n}]$.

Repeat as needed, to build a sequence $\{R_i\}_{i=1}^\infty$ Show that it converges to an accumulation point.