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.
@"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.