Textbook has wrong answer? - Metric spaces "topological properties" (probably trivial for the confident) In the book there's a table and above it it reads "we have crossed out the wrong answer" meaning the remaining one is right. I dispute this, there are 4, I thought I got the first one right, but I don't have the confidence to assert they are all the wrong way around.
The question is "for the following decide which are a topological property"
1) True is crossed out - meaning the following is false
$(\forall x\in M)(\exists y\in M)d(x,y)=1$
I have interpreted this as "for every x in the space there's a y in the space such that the metric between them is 1" - I agree this is false, I have seen no reason/definition that a metric must take the value 1 at all. 
2) False is crossed out - meaning the following is true
$(\forall x\in M)(\exists y\in M)d(x,y)=-1$
I have interpreted it the same way, and I believe that the metric is defined to be $\ge0$ specifically $0\iff x=y$ - So I dispute this "answer"
3) True is crossed out meaning the following is false
$(\forall x\in M)(\forall e>0)(\exists y\in M)d(x,y)>e$
I agree it is false, as we could choose a huge $e$ with a metric that only takes the values 0 and 1 - so I agree
4) False is crossed out - meaning this is true
$(\forall x\in M)(\forall e>0)(\exists y\in M)0<d(x,y)<e$
I agree as a topological space consists only of open sets, which means there is a neighborhood/ball around each point. 
It really is that second one, I do not believe that is true. It has damaged my confidence in my reasoning for the others.
 A: You are not being asked to decide whether the statements are true or false, but rather whether they are topological properties: presumably this means that the statement holds for a given metric iff it holds for any equivalent metric (i..e, which determines the same topology).
Since the second statement never holds for a metric, it is true that it holds for a metric iff it holds for any equivalent metric.  (It's sort of a silly question...)
A: Thanks for positing your other answers, and the reasoning. I think that you may be right to question your general approach here.  
As you know, given a metric space $\mathcal{X}=\langle X,d\rangle$, there is a topological space $\mathcal{X}'=\langle X,\tau\rangle$, where $\tau$ is the set of $d$-open sets in $\mathcal{X}$ (although the converse is not true - not all topologies admit metrics!). 
You can think of a topological property of a $\mathcal{X}$ is one that can be defined in terms of this $\tau$, without explicit reference to $d$.  The formal definition is that $T$ is a topological property if for any two metric spaces $\mathcal{X}$ and $\mathcal{Y}$ for which there is a homeomorphism $f:\mathcal{X}\cong\mathcal{Y}$, we have 
$$T(\mathcal{X})\iff T(\mathcal{Y}).$$
(A homeomorphism is an invertible continuous function whose inverse is also continuous.  Homeomorphisms form the isomorphisms in the category $\textbf{Top}$ of topological spaces.  Note that being a homeomorphism is a strictly weaker property than being an isometry (isometries are the isomorphisms in $\textbf{Met}$): all isometries are homeomorphisms but not all homeomorphisms are isometries). 
A special case of this kind of thing comes up when you consider equivalent metrics: recall that two metrics $d$ and $d'$ on the same set $X$ are equivalent if the open sets induced by $d$ are the same as the open sets induced by $d'$.  In this case, the homeomorphism we are looking for is the identity map $X\to X$. 
Anyway, the question is asking which properties are topological, in the sense that they are definable in terms of the topology of the metric spaces.  


*

*In the case of the first statement, the point is that you can have two topologically equivalent metric spaces, one of which has lots of pairs of points separated by distance 1, and one of which does not.  As a simple example, consider the set $X=\{a,b\}$, let $d_{1}$ be the metric on $X$ given by $d_1(a,b)=1$ and let $d_{2}$ be the metric on $X$ given by $d_2(a,b)=2$.  Then the identity map $$\textrm{id}_{X}:\langle X,d_1\rangle\to\langle X,d_{2}\rangle$$ 
is a homeomorphism.  But $\langle X,d_1\rangle$ satisfies the stated condition while $\langle X,d_2\rangle$ does not.  The property is therefore not a topological property: its truth or falsity depends on non-topological features of the metric. 

*As @Pete L. Clark noted in his answer, it is vacuously true that the second property is a topological property: the stated property is false for all metric spaces, and so the biconditional statement $T(\mathcal{X})\iff T(\mathcal{Y})$ trivially holds for any $\mathcal{X}\cong\mathcal{Y}$.

*The third property is essentially equivalent to "$\mathcal{X}$ is unbounded."  You are right that this is not a topological property.  The easiest way to do this is to note that, with the usual metrics, $(0,1)$ and $(-\infty,\infty)$ are topologically equivalent - there are plenty of homeomorphisms witnessing this fact.  But $(0,1)$ is bounded while $(-\infty,\infty)$ is not.  Thus, boundedness is not a topological property (unlike compactness, by the way, which is a topological property).

*For the last example, you are again correct: the property is a topological property.  But the reason why is a little harder to prove.  I'll leave you to work out the details, but you might start with a single set $X$ and two equivalent metrics $d_{1}$ and $d_{2}$  Assume that $\langle X,d_{1}\rangle$ has the desired property, and prove that $\langle X,d_{2}\rangle$ also has it.  As a hint, consider the characterization of equivalent metrics saying that $d_{1}$ and $d_{2}$ are equivalent if and only if for every $x\in X$ and every $\epsilon_{1}>0$, there is an $\epsilon_{2}>0$ such that 
$$B_{d_{2}}(x,\epsilon_{2})\subseteq B_{d_{1}}(x,\epsilon_1).$$ (The fact that the property in question is so easily phrased in terms of neighborhood filters is a dead giveaway that it is a topological property.)
