Show that topologies are the same I just read a proof where it was said that if for each element in the topology 2 we find an element in topology 1 that is contained in this set and vice versa, then they are the same. How do I see that this is equivalent to the notion that the topologies are subsets of each other? By the way, one of them was induced by a metric.
 A: As stated, the observation made is incorrect, as noted in the comment by drhab above:

For each element in topology 2 you can always find an element in topology 1 that is contained in this set: the empty set. And vice versa off course. So in general this cannot be true. Spread some light over this.

Even without being quite so facetious it is false: Let $\mathcal{O}_{\text{m}}$ denote the usual (metric) topology on $\mathbb{R}$, and let $\mathcal{O}_{\text{S}}$ denote the Sorgenfrey (lower-limit) topology on $\mathbb{R}$.

*

*Given any (nonempty) $U \in \mathcal{O}_{\text{m}}$ we can find $a < b$ such that $(a,b) \subseteq U$, and so $[ \frac{ a+b}{2} , b ) \subseteq (a,b) \subseteq U$, and $[ \frac{ a+b}{2} , b ) \in \mathcal{O}_{\text{S}}$.


*Given any (nonempty) $V \in \mathcal{O}_{\text{S}}$ we can find $a < b$ such that $[a,b) \subseteq V$, and so $(a,b) \subseteq [a,b) \subseteq V$ and $(a,b) \in \mathcal{O}_{\text{m}}$.

The correct statement would be the following:

Suppose that $\mathcal{O}_1$ and $\mathcal{O}_2$ are two topologies on a set $X$.  Then $\mathcal{O}_1$ and $\mathcal{O}_1$ are the same iff the following two things hold:

*

*Given any $U \in \mathcal{O}_1$ and any $x \in U$ there is a $V \in \mathcal{O}_2$ such that $x \in V \subseteq U$; and

*Given any $V \in \mathcal{O}_2$ and any $x \in V$ there is a $U \in \mathcal{O}_1$ such that $x \in U \subseteq V$.


The analogous statement is also true for bases for topologies on $X$:

Suppose that $\mathcal{B}_1$ and $\mathcal{B}_2$ are two bases for topologies on a set $X$.  Then the topologies generated by $\mathcal{B}_1$ and $\mathcal{B}_1$ coincide iff the following two things hold:

*

*Given any $U \in \mathcal{B}_1$ and any $x \in U$ there is a $V \in \mathcal{B}_2$ such that $x \in V \subseteq U$; and

*Given any $V \in \mathcal{B}_2$ and any $x \in V$ there is a $U \in \mathcal{B}_1$ such that $x \in U \subseteq V$.



Added:
Consider the following slightly modified version of the given statement:

For each nonempty $U \subseteq X$ open with respect to the first topology, there is a nonempty $V \subseteq Y$ open with respect to the second topology such that $V \subseteq U$, and vice versa.

What this statement says is that every (nonempty) open set in the first topology has nonempty interior with respect to the second topology (and vice versa).
It should be noted that two topologies on a set $X$ sharing this property may not even be comparable (let alone the same).  The simplest example I can think of are the following topologies on $\mathbb{R}$: the Sorgenfrey (lower-limit) topology generated by $\{ [a,b) : a < b \}$ and the "upper-limit" topology, generated by $\{ ( a , b] : a < b \}$.  (Of note is that while these topologies are not comparable, they are homeomorphic via the mapping $x \mapsto -x$.)
