If we omit condition (2) in the definition of the directed set, then it is the same as to assume that there could be classes of different but "equivalent" indexes. Therefore, this does not restrict the previous definition, and all the examples in the Exercise 1 are still nets. The exercise 2 showed that a cofinal subset of a directed set is a directed set itself, and the proof remains valid as it does not depend on the property (2). The Exercise 3 shows that a sequence is a net, and since the new definition of a net enlarges the set of all nets, the fact still holds. In the exercise (4) we showed that a pointwise convergence of a finite product of nets implies the convergence of the product and, once again, the proof does not depend on the property (2). Exercise 5 tells us that in a Hausdorff space a net cannot converge to two different points at the same time, and the result is still valid as the proof depends on the fact for a given pair of elements there is another one greater than both whether the two are equivalent or not.Now, we claim that for any net $(x_\alpha)_{\alpha\in J}$ satisfying the new definition there is a subnet that is also a net according to the initial definition. Indeed, consider the new index set $[J]$ that consists of all equivalence classes $[\alpha]$ of $J$ with the partial order given by the previous relation (i.e. $[\alpha]\preceq[\beta]$ iff $\alpha\preceq\beta$ ). It is straightforward to see that this new relation is well-defined and is a (strict) partial order on the set of all equivalence classes. Take the map $f:[J]\to J$ by taking $f([\alpha])$ to be any element from $[\alpha]$ . This map preserves the relation, and its image is cofinal in $J$ as $f([\alpha])\succeq\alpha$ for all $\alpha$ . Let us call a subnet constructed this way $(x_{[\alpha]})$ (there may be many as the choice of $f([\alpha])$ is arbitrary).If $(x_\alpha)$ converges to some point $x$ , then $(x_{[\alpha]})$ converges to the same point. In fact, this holds more generally, namely, if a net converges to a point, then every its subnet converges to the same point. This was proved in Exercise 8, and the proof remains true for the new definition as well. Using this and the construction of a subnet satisfying the initial definition given above, we conclude that the theorem of Exercise 6 (a point is in the closure of a set iff there is a net in the set converging to the point) still holds.For Exercise 7 we need a more general result. Suppose that for a net $(x_\alpha)$ that satisfies the new definition every its subnet that satisfies the initial definition converges to $x$ . Then $x_\alpha\to x$ . [ Suppose not. Then there is a neighborhood $U$ of $x$ such that for every $\alpha$ there is $\gamma(\alpha)\succeq\alpha$ such that $x_{\gamma(\alpha)}\notin U$ . For each $[\alpha]$ define $f([\alpha])$ to be an element $\beta\in[\alpha]$ such that $x_\beta\notin U$ if such an element exists or any arbitrary element in $[\alpha]$ otherwise. Then this defines a subnet of the type $x_{[\alpha]}$ the same way as before, i.e. it satisfies the initial definition. For any $[\alpha]$ : $x_{f([\gamma(\alpha)])}\notin U$ and $[\alpha]\preceq[\gamma(\alpha)]$ . This contradicts the assumption. ] Given this result, we conclude that if $f$ is continuous and a net (in the new definition) converges to $x$ then every its subnet satisfying the initial definition converges to $x$ and its image converges to $f(x)$ , therefore, the image of the net converges to $f(x)$ as well. The reverse implication is immediate.The lemma of Exercise 9 tells us that $x$ is an accumulation point of the net $(x_\alpha)$ iff its a limit of a subnet. Note, that $x$ is an accumulation point of $(x_\alpha)$ iff for any neighborhood $U$ of $x$ the set of all indexes $\alpha$ such that $x_\alpha\in U$ is cofinal in $J$ iff the set of all indexes $[\alpha]$ such that there is $\beta\in[\alpha]$ such that $x_\beta\in U$ is cofinal in $[J]$ iff there is a subnet $(x_{[\alpha]})$ that has an accumulation point $x$ . This implies that there is a subnet converging to $x$ . The other direction: if there is a subnet $x_{g(\beta)}$ converging to $x$ then we can leave the proof that it is an accumulation point the same as before.For the Exercise 10 suppose, first, that every net has a convergent subnet. Then every net according to the initial definition has a convergent subnet that has a convergent subnet satisfying the initial definition. Therefore, as was proved before, the space is compact. Now, suppose that the space is compact. Then for a net there is a subnet satisfying the initial definition that has another subnet converging to a point.Finally, the proof of the Exercise 11 does not depend on the definition of the net but rather on the previous results which we have already proved to hold under the new definition.