Munkres states that equicontinuity depends on metrics

[sammycaps]

So munkres states that equicontinuity depends on the metric and not only on the topology. I'm a little confused by this. Is he saying that if we take C(X,Y) where the topology on Y can be generated by metrics d and p, then a set of functions F might be equicontinuous in one and not the other? This seems unlikely as continuity is a topological property, and equicontinuity just seems to be a bit of a tweak on this for function spaces.

 

[HallsofIvy]

Yes, continuity is a topological property and the topology depends upon the metric. Different metrics, different topologies, and different continuous functions.

For example, suppose X and Y are the set of real numbers with usual metric and so the usual continuous functions. Now take p to be the "discrete" metric, p(x,y)= 1 if x\ne y, p(x, x)= 0. In this metric all sets are open so that, for example, the function f(x)= x, which is continuous in the usual metric, is not continuous. The inverse image of the open set (in the disrete metric) {1} has inverse image {1} which is NOT open.

 

[sammycaps]

Thank you, but that is not quite my question. I know that different metrics may induce different topologies. But, some properties are properties of the metric and not the topology (like boundedness), as in two metrics can give the same topology but the space may be bounded in one and not the other (i.e. standard bounded metric).

It seems clear that equicontinuity is a property of the topology (which, of course, changes with different metrics), rather than a property purely of the metric (like boundedness). However, the way Munkres phrased it, I just want to be sure I'm not missing some subtlety.

 

[micromass]

Yes, continuity is a topological property and the topology depends upon the metric. Different metrics, different topologies, and different continuous functions.

For example, suppose X and Y are the set of real numbers with usual metric and so the usual continuous functions. Now take p to be the "discrete" metric, p(x,y)= 1 if x\ne y, p(x, x)= 0. In this metric all sets are open so that, for example, the function f(x)= x, which is continuous in the usual metric, is not continuous. The inverse image of the open set (in the disrete metric) {1} has inverse image {1} which is NOT open.

I don't think that's what he meant to ask. He meant to ask if there are two metrics which generate the same topology, but for which a family is equicontinuous in the first metric but not in the second metric.

If two metrics generate the same topology, then a function is continuous in the first metric if and only if it is continuous in the second.

But anyway, I think Munkres is wrong. You can define equicontinuity in just topological spaces.

Let $X$ and $Y$ be topological spaces. Let $A$ be a set of continuous functions between $X$ and $Y$. This set is said to be equicontinuous if for all $x\in X$ and $y\in Y$ and if for all open set $G\subseteq Y$ that contains $y$, there exists a neighborhood $U$ of $x$ and a neighborhood $V$ of $Y$ such that for each $f\in A$ holds that if $f(U)\cap V\neq \emptyset$ then $f(U)\subseteq G$.

However, this is not the only way to generalize equicontinuity. Another way is

Let $X$ and $Y$ be topological spaces. Let $A$ be a set of continuous functions between $X$ and $Y$. This set is said to be equicontinuous if for all $x\in X$ and $y\in Y$ and if for all open set $G\subseteq Y$ that contains $y$, there exists a neighborhood $U$ of $x$ and a neighborhood $V$ of $Y$ such that for each $f\in A$ holds that if $f(x)\in V$ then $f(U)\subseteq G$.

These two definitions are equivalent to each other (and to the normal definition of equicontinuity) in metric spaces. But they are no longer equivalent in metric spaces (although the first definition always implies the second).

Of course, I guess you can also define equicontinuity as

Let $X$ and $Y$ be topological spaces. Let $A$ be a set of continuous functions between $X$ and $Y$. This set is said to be equicontinuous if for all $x\in X$ and if for all open set $G\subseteq Y$ that contains $f(x)$, there exists a neighborhood $U$ of $x$ such that for all $f\in A$ holds that $f(U)\subseteq G$.

But they don't seem to work with this definition in topological spaces, I have no idea why not.