Mathematics Asked by Jpmarulandas on December 12, 2020

While studing from Nakahara’s book of Differential Geometry, I reached the part where topological spaces are introduced, and to define a topological space $mathcal{T}$, they take it as a collection of subsets of a given set $X$, such that $mathcal{T} = left{ U_i | i in I right}$. This may be a stupid question, but I have not been able to figure what does the set $I$ is, to which $i$ belongs to. I have thought it to be the set of all possible subsets of $X$, or the set that contains the element indexes of the subsets of $X$, but these do not sound correct to me at all.

A topology $mathcal T$ on a set $X$ is a set of subsets of $X$ satisfying the well-known axioms for "open sets". Thus, letting $mathfrak P(X)$ denote the powerset of $X$, we have $mathcal T subset mathfrak P(X)$.

In my opininon there is no benefit to regard a topology as an *indexed collection* ${U_i mid iin I}$. Writing it in this form means that we are given a set $I$ of indices and for each $i in I$ an element $U_i in mathfrak P(X)$. The index set $I$ can be *any* set. You should be aware that such an indexed collection of sets $U_i$ formally is a *function* $phi : I to mathfrak P(X)$ where we write $U_i = phi(i)$. Note that this allows $U_i = U_{i'}$ for $i ne i'$, i.e. your indexed family of sets $U_i$ may have multiple occurences of the same set.

If we simply say that $mathcal T subset mathfrak P(X)$, then we can regard $mathcal T $ as indexed collection by taking $I = mathcal T$ and letting $phi : mathcal T to mathfrak P(X)$ be the inclusion, i.e. $phi(U) = U$. This may be called the "self-indexing" of $mathcal T$.

Anyway, both approaches are correct, thus it is a matter of taste which you prefer.

Answered by Paul Frost on December 12, 2020

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