Can we think of a sequence as of point in space?

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At today's real analysis lecture our professor has told us, that sequences are ordered collection of objects. After lecture a question have bumped in my head - can we think of sequence $(a_n)_{n=1}^{n}$ as of point in $n$-dimensional space?

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Can be used in a sense to construct the real numbers. Equivalence classes of convergent Cauchy sequences. Pretty interesting. You will learn about it later in analysis.

EDIT: they way you are defining it is essentially an n-tuple, which is a point in $\mathbb{R^n}$: ($a_1$,...,$a_n$)

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Absolutely. Assuming your $a_n$ are real, the sequence gives you coordinates so you have a point in $\Bbb R^n$. If they come from some other set, the sequence gives you a point in the $n$-fold Cartesian product of that set.

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Sure, and it is not true only for finite dimensional spaces. Given a set $X$ we can define $$X^{\mathbb{N}} = \{(x_1,\ldots,x_n,\ldots): x_i \in X\}.$$ It is possible to show that if $X$ is a vector space, so is $X^{\mathbb{N}}$.

In a more formal context, a sequence $X$ is defined to be a function from $\mathbb{N}$ to $X$. Consequently, your each point of your set $X^{\mathbb{N}}$ is a function.

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Yes. The reason is because there is an isomorphism between the vector space of all n-tuples of $\mathbb R$ and the set of all sequences of n elements of $\mathbb R$, both under pointwise addition and scalar multiplication.

You can, and in most classes do, use n-tuples and sequences interchangeably by associating with each sequence the n-tuple whose ith element is the ith element of the sequence and each n-tuple, the sequence whose ith element is the ith element of the n-tuple.

If switching all the time changes you, you can just choose one and stick to it as well, although I recommend being able to understand it well enough that you can think about the theory you're learning both ways, for sequences and for n-tuples. Notice also that you can graph a sequence in n-space in exactly the same way you'd graph an n-tuple.

Adam V. Nease