Decide whether the following properties of metric spaces are topological or not I have to decide whether the following properties of metric spaces are 'topological' or not,
(a) every continuous function on M is bounded
(b) $(\forall x \in M)(\forall y \in M) d(x; y) > 0$
(c) $(\forall x \in M)(\forall y \in M) d(x; y) > 1$
Where M is a metric space with metric d. 
but looking through my notes.. i can't find any distinct definition of what 'topological' means. Would anyone be able to clear this up?
 A: A topological property is any property that is preserved by all homeomorphisms.  A homeomorphism from one metric space $X$ to another, $Y$, is a function $f:X\to Y$ such that


*

*$f$ is continuous;

*$f$ is one-to-one;

*$f^{-1}$ is continuous;

*$f$ is onto, i.e. every point in $Y$ is the image of some point in $X$.


Suppose one defines a "purple" metric space.  To say "purpleness" is a "topological property" means that for every metric space $Y$ for which a homeomorphism from $X$ to $Y$ exists is also purple.
So your task would be: either


*

*Show that whenever a function $f$ exists with the properties above and $X$ is "purple", then the space $Y$ must also be "purple", then conclude that "purpleness" is a topological property of metric spaces; or

*Show that some metric space $Y$ and some function $f$ with the properties above exists and that $X$ is "purple" and $Y$ is not "purple", then conclude that "purpleness" is not a topological property.


So let's consider your first question:


*

*If every continuous function $g:X\to\mathbb R$ is bounded and $f:X\to Y$ is a homeomorphism from the metric space $X$ to some other metric space $Y$, is that enough to prove that every continuous function $h:Y\to\mathbb R$ is bounded?  If so, then the property involved is topological.

*But if there is some metric space $Y$ and some function $f$ with the propertires above, and some function $h:Y\to\mathbb R$ that is not bounded, then the property involved is not topological.

A: I would expect that a property could be called 'topological' if it can be reformulated without any appeal to the metric.
However, first you have to deal with the fact that your second two properties are always false. Maybe you should check your source.
Since no response of the OP seems to be forthcoming, I will answer the questions that I suspect should have been actually asked:

a) Every continuous function from $M$ to any other metric space $X$ is bounded.

Yes, it is a topological property:
Being continuous can be expressed entirely in terms of the topology, for example with the open sets of the space. Therefore, a continuous function will be continuous no matter which metric is chosen to generate the same topology. So clearly, the question whether all functions in this set are bounded only depends on the topology, but not on the metric.
(The version in the OP is somewhat unclear what functions are to be regarded. If we look at functions from somewhere else to $M$ or from $M$ to $M$, the answer will change, because then the boundedness will be defined in terms of the metric of $M$. However, I suspect that the above is the intended interpretation because of its relation to compactness.)

b) $d(x,y)>0$ for all $x\not=y$ in the given metric space.

Yes, it is a topological property, but not a very interesting one:
In any metric $d(x,y) >0$ is true for any $x\not= y$, so this property is always true and certainly does not depend on the choice of metric.
(The original version in the OP is trivally always false, of course, and therefore also a topological property.)

c) $d(x,y)>1$ for all $x\not=y$ in the given metric space.

No, this is not a topological property, it depends on the metric.
Take the discrete space with two points and set their distance to 1. This metric space clearly does not fulfill the property. If you instead choose the metric where this distance is 2, this clearly gives you the same topology, but now the property is fulfilled. Therefore, the property depends on the choice of metric, it is not a topological property.
(The original version in the OP is trivally always false, of course, and therefore a topological property.)
A: In short, a property of a metric space is topological if it does not (really) depend on the metric. That is, if you have a different metric that produces the same notions of convergence, continuous functions, etc. (or even if you only have the notions of convergence, contnuou functions, etc.) then the property does not change.
Note that $b$ and $c$ can be greatly simplified to a truth value that holds completely independently of the metric $d$; which makes at least these properies topological. 
A: @"user65972" : Maybe you're hesitating about what to think of the answers, or maybe you've just been distracted by other demands of your life.  Since nothing's being done about the weird situation in the comments (I've never actually "disputed", as opposed to merely corrected, an answer before . . . . .) I feel moved to post an answer with a different point of view, which perhaps you should not read unless you can make no headway based on my earlier answer.
Suppose $X$ is a metric space and every continuous function on $X$ is bounded.  I'll assume we're talking about continuous functions from $X$ into some other metric space, not necessarily real-valued functions.
Is that a topological property?
That question can be answered as follows: Suppose $Y$ is a metric space homeomorphic to $X$; that means there is a homeomorphism $f:X\to Y$.  In other words, $f$ is one-to-one and onto and $f$ and $f^{-1}$ are continuous.  Does it necessarily follow that every continuous function on $Y$ is bounded?  If so, then the property we're considering is a topological property.
So suppose $h$ is a continuous function whose domain is $Y$.  How can we show that $h$ must be bounded?  The proof must rely on the existence of the homeomorphism $f$.  We have
$$
X \overset{f}{\longrightarrow} Y \overset{h}{\longrightarrow} \text{some other space}.
$$
Then the function $h\circ f:X\to\text{some other space}$ is continuous, hence bounded.  That means the set
$$
\{h(f(x)) : x\in X\} \tag 1
$$
is bounded.  But $f$ is onto the space $Y$, so every point $y\in Y$ is $f(x)$ for some $x\in X$.  We can deduce that the set $(1)$ is the same as the set
$$
\{h(y) : y\in Y\}. \tag 2
$$
(If that's not immediately evident, then think about how to show that $(1)$ and $(2)$ are subsets of each other.)  The set $(2)$ is therefore bounded.  Since $h$ was an arbitrary continuous function on $Y$, it follows that EVERY continuous function on $Y$ is bounded.
Therefore this is a topological property.
An example of how to show something is NOT a topological property is this: The space $(0,1)$ with the usual metric is bounded, and the space $(-\infty,\infty)$ with the usual metric is unbounded.  But
$$
f(x) = \frac{1}{1+2^x}
$$
is a homeomorphism from $(-\infty,\infty)$ to $(0,1)$.  Since the two spaces are therefore homeomorphic to each other, but one is bounded and the other is not, we conclude that boundedness of metric spaces is NOT a topological property.
