# Why the neighborhood is a topological space?

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)?

• 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 $\{f[t]:t\in T_X\}=T_Y.$ Feb 1, 2020 at 10:10

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).

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).

• thanks - I need this answer Feb 1, 2020 at 3:43