Question: Suppose $f(x)$ is periodic with period p and $g(x)$ is periodic with period q. Let $r$ be the L.C.M. of p and q, if it exists. Then show that:

1. If $f(x)$ and $g(x)$ cannot be interchanged by adding a least positive number $k<r$, to $x$, then $r$ is the period of $f(x)+g(x)$.
2. If $f(x)$ and $g(x)$ can be interchanged by adding a least positive number $k<r$, to $x$, then $k$ is the period of $f(x)+g(x)$.

I know I should give my approach towards the problem but I can't seem to be able to make the head or tail of this problem. I don't think I even completely understand what's being said.

Answer: 
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.
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.)