In M.A. Armstrong's Basic Topology, he introduces the concept of a topological space by first defining it in terms of neighborhoods (rather than open sets). Here is his formulation (p. 13):
We ask for a set $X$ and for each point $x$ in $X$ a nonempty collection of subsets of $X$, called neighborhoods of $x$. These neighborhoods are required to satisfy four axioms:
(a) $x$ lies in each of its neighborhoods
(b) The intersection of two neighborhoods of $x$ is itself a neighborhood of $x$.
(c) If $N$ is a neighborhood of $x$ and if $U$ is a subset of $X$ which contains $N$, then $U$ is a neighborhood of $x$.
(d) If $N$ is a neighborhood of x and if $\mathring{N}$ denotes the set $\{z\in N | N$ is a neighborhood of $z\}$, then $\mathring{N}$ is a neighborhood of $x$.
This whole structure is called a topological space.
He then defines the subspace topology as follows:
Let $X$ be a topological space and let Y be a subset of X. We can define a subspace topology on Y as follows. Given a point $y\in Y$ take the collection of its neighborhoods in the topological space $X$ and intersect each of these neighborhoods with $Y$. The resulting sets are the neighbourhoods of $y$ in $Y$.
One of the exercises in this section is to veryify that the subspace topology (as defined above) satifies axioms (a)-(d) and is, hence, a topology on Y. I was able to verify axioms (a)-(c) fairly easily; however, I can't seem to finish the verification of axiom (d).
This is what I have so far:
To verify axiom (d), suppose $y\in Y$, $N_Y$ is a neighborhood of $y$ in $Y$, and let $$\mathring{N_Y}=\{z\in N_Y | N_Y \text{ is a neighborhood of } z\}.$$ Observe that $\mathring{N_Y}\subset N_Y \subset Y$ and, by definition, $N_Y=Y\cap N_X$ for some neighborhood $N_X$ of $y$ in $X$. Now let $$\mathring{N_X}=\{z\in N_X | N_X \text{is a neighborhood of }z\}.$$ By axiom (d), $\mathring{N_X}$ is a neighborhood of $y$ in $X$.
We will show that $\mathring{N_Y}=Y\cap \mathring{N_X}$, which will show that $\mathring{N_Y}$ is a neighborhood of $y$ in $Y$.
I was able to prove that $(Y\cap \mathring{N_X}) \subset \mathring{N_Y}$, but I have not been able to establish that $\mathring{N_Y} \subset (Y\cap \mathring{N_X})$. If I take an arbitrary $z\in \mathring{N_Y}$, then I know that $N_Y=Y\cap N_X$ is a neighborhood of $z$ in $Y$, but I do not see how to deduce from this that $N_X$ is a neighborhood of $z$ in $X$ (and hence that $z\in \mathring{N_X}$).
I feel like there is just something obvious that I am missing, or, perhaps, my intuition was wrong and $\mathring{N_Y}\neq Y\cap \mathring{N_X}$, and axiom (d) is verified in some other way.
Any hints/suggestions would be much appreciated. Thanks.
