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I'm a total dilettante in topology, thus, my apologies for this question. Wikipedia writes that:

"...each point of an n-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension n."

As I understand, a homeomorphism is a sort of isomorphism between topological spaces. So, the neighbourhood must be a topological space. What this space is (i.e. its open sets, or base)?

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    $\begingroup$ A topology on a set $X$ is a special kind of collection of some or all of the subsets of $X$....When $t$ is a subset of the domain of a function $f,$ we write $f[t]=\{f(x): x\in t\}$.... If $T_X$ is a topology on $X$ and if $T_Y$ is a topology on $Y$ then a homeomorphism from $(X,T_X)$ to $(Y,T_Y)$ is a bijection $f:X\to Y$ such that $\{f[t]:t\in T_X\}=T_Y.$ $\endgroup$ Feb 1, 2020 at 10:10

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A neighborhood is an open set, i.e. a subset of your space (manifold). Thus, the topology on this subset should be the subset topology induced by the manifold. That is so say, it is a topological space with the topology induced by your whole space (manifold).

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If $(X,\tau)$ is a topological space and $Y\subseteq X$, there is a natural way to put a topology on $Y$ which "comes from" $\tau$: namely, we let $$\tau[Y]=\{U\cap Y: U\in\tau\}.$$ This is called the subspace topology on $Y$, and in this way all subsets of (the set of points of) a topological space can be viewed as topological spaces themselves in a "canonical" way.

There are a few points worth making:

  • This works for any $Y\subseteq X$, not just the closed, open, or otherwise "nice" subsets.

  • The subspace $(Y,\tau[Y])$ may be very different from the original space $(X,\tau)$. A standard set of exercises is to check how various properties are or are not retained: if $(X,\tau)$ is Hausdorff, must $(Y,\tau[Y])$ also be Hausdorff? If $(X,\tau)$ is compact, must $(Y,\tau[Y])$ also be compact? And so forth.

  • Note that the subspace topology is not given by the set of open-in-the-original-sense subsets of $Y$: $\{U: U\subseteq Y, U\in \tau\}$ is very different from $\{U\cap Y: U\in\tau\}$, and indeed the former won't be a topology in general (it isn't guaranteed to contain $Y$ itself).


A good exercise about the subspace topology is to check that if we take $\mathbb{R}^2$ with the usual topology, the subspace topology on the $x$-axis is homeomorphic to the usual topology on $\mathbb{R}$. More generally, subspace topologies tend to be pretty easy to visualize; they more-or-less comport with our expectations (compare with product and quotient topologies, which have more subtleties early on).

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  • $\begingroup$ thanks - I need this answer $\endgroup$ Feb 1, 2020 at 3:43

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