Solution to exercise seems to contradict interchangeability of open and closed sets

Question

I'm working on an exercise from baby rudin, where the question is

If $f$ is a real continuous function defined on a closed set $E\subset\mathbb{R}^1$, prove that there exist continuous real functions $g$ on $\mathbb{R}^1$ s.t. $g(x)=f(x)$ $\forall x\in E$. We call functions such as $g$ continuous extensions of $f$ from $E$ to $\mathbb{R}^1$. Show that the result becomes false if the word "closed" is omitted. Extend the result to vector-valued functions.

In my solution is as follows

For $x$ on the boundary of $E$, every neighbourhood $U$ of $x$ is mapped to a neighbourhood of $f(x)$. Since $g(x)=f(x)\;\forall x\in E$, every neighbourhood of $x$ is also mapped to a neighbourhood of $g(x)$. As such, the open set $U\cap E^c$ is also mapped to an open set, and since this is true for every neighbourhood $g(x)$ is continuous on $\mathbb{R}$. Since we have used topological properties only, this result holds for any closed set $E$ in any topological space.

When $E$ is not closed, we can proceed simply by counter-example. Suppose $E$ is an open set. We define $g(x)\ne f(x)$ when $x$ is on the boundary of $E$. Then $f(x)$ and $g(x)$ are continuous inside of $E$, but $g$ has a discontinuity along the boundary of $E$. The discontinuity can always exist because every point in $E$ has a neighbourhood which is a proper subset of $E$, so every subset where $f(x)=g(x)$ can be separated from the neighbourhood of a point on the boundary of $E$.

However, from my understanding of topology, the open and closed sets can be used interchangeably, i.e. we can define the elements of a topology to be "closed sets", and so the complements of the elements of the topology would be our "open sets". However, since I've used the limit point definition of a closed set, this seems to contradict this topological behaviour, since from my proof the continuous extension is not possible for sets which do not contain all their limit points. This implies to me that if I were to define the topology using closed sets, the result of the exercise would not hold. However, in my proof I've used the open set definition of continuity, which is the one used in topology. What have I misunderstood which has caused this discrepancy?

Answer 1

The open and closed sets are certainly not interchangeable. The set $S$ of all closed subsets of $\Bbb R$ is not a topology. For instance, the union of all $[0,1-1/n]$ is the set $[0,1)$ which is not in $S$, so $S$ is not even closed under countable unions (never mind arbitrary unions).

Your solution is not a solution. It presupposes the existence of an (undefined) $g$ but the goal is to show there is a continuous extension $g$. The claim is not true for all topological spaces, so that is also a wrong conclusion; the general Tietze extension theorem works for all normal (not necessarily Hausdorff) spaces but it doesn't work for all possible spaces.

A simple counterexample; take any infinite cofinite space. The only continuous real valued functions thereon are the constant functions. However, all finite subspaces are (a) closed and (b) discrete so you can easily define non-constant continuous functions on them; they cannot possibly extend to the whole space.

I note your counter-example is not a counter-example; you claim without justification that you "by defining $g(x)\neq f(x)$" (this is not a definition) you can get a continuous $g$. This is without justification. Moreover, even if you could justify it it would be insufficient; you have to show no such $g$ exists, not just that "there is a $g$ which doesn't extend $f$" (which is always true).

It's also not true that if $E$ is open then no such extension can exist. Take any space, any continuous real valued function on that space; by restricting it to any open subspace, you get a map which does have a continuous extension (the original function...).