Topology: Can We Use Same Function for 2 Open Sets?

[kent davidge]

In reading out about topological spaces and topologies I noticed that they do not give much specific examples, so I have not found an answer to the following simple question:

Can we use the same function for mapping into two different open sets of a given topology? Or, perhaps equivalently, can we pick as the domain of a function, the union of two open sets of a given topology?

 

[FactChecker]

Maybe I misunderstand your question. What kind of function are you talking about? And how are your last two sentences equivalent when the first one is about open sets in the range and the other is about open sets in the domain?

 

[fresh_42]

In reading out about topological spaces and topologies I noticed that they do not give much specific examples, so I have not found an answer to the following simple question:

The origin and a standard example for topological spaces are metric spaces like $\mathbb{R}^n$. Although topological spaces are far more general than this, they remain examples. It's just that we can define open sets more generally than induced by a metric and have still many tools like continuity left. The generalization is basically that open balls $|x-x_0|<\varepsilon$ turn into open sets.

Can we use the same function for mapping into two different open sets of a given topology?

Given $f\, : \,\mathbb{R} \longrightarrow \mathbb{R}$, say $f(x)=\sin(x)$ can we use $f(x) \in (-2,2)$ and $f(x)\in (-5,5)\,$? Does this make sense? However, we can define $\sin x\, : \,\mathbb{R} \longrightarrow [-1,1]$ where the codomain contains only the interval which is thus open. It's called subspace topology and contains all sets $\mathcal{O} \cap [-1,1]$ with $\mathcal{O} \subseteq \mathbb{R}$ open as open sets in $[-1,1]$.

Or, perhaps equivalently, can we pick as the domain of a function, the union of two open sets of a given topology?

We usually pick a function and determine its domain and not the other way around. Your question sounds like: Can we pick a function $f\, : \,\mathbb{R} \longrightarrow (a,b) \cup (c,d)\,$? Hmm, why not? But what for?

 

[WWGD]

In reading out about topological spaces and topologies I noticed that they do not give much specific examples, so I have not found an answer to the following simple question:

Can we use the same function for mapping into two different open sets of a given topology? Or, perhaps equivalently, can we pick as the domain of a function, the union of two open sets of a given topology?

If I understood correctly, answer to first is yes: as long as the output is contained in the union of the open sets. For the second, yes, as long as the function is defined in both the sets. For the second, In some cases you may also want to preserve some properties of your function ( continuity, compactness, etc. or non-topological ones like differentiability), which may restrict the open sets over which you can define your function. This is called the extension problem and it can become pretty tricky.

 

[kent davidge]

If I understood correctly, answer to first is yes: as long as the output is contained in the union of the open sets. For the second, yes, as long as the function is defined in both the sets. For the second, In some cases you may also want to preserve some properties of your function ( continuity, compactness, etc. or non-topological ones like differentiability), which may restrict the open sets over which you can define your function. This is called the extension problem and it can become pretty tricky.

So what's the difference if we take, say, the trivial topology and a function from the set to itself and another topology and the same function from the set to itself? I thought what makes it different was the open sets in which the function will act on.

 

[fresh_42]

So what's the difference if we take, say, the trivial topology and a function from the set to itself and another topology and the same function from the set to itself? I thought what makes it different was the open sets in which the function will act on.

What makes it different is the determination of the topology, i.e. the selection of open sets. This has consequences for the property of functions, not the functions itself. Take for example
$$
f(x)=\begin{cases} 0 &\text{ for } x \leq 0 \\ 1 &\text{ for } x > 0 \end{cases}
$$
which is not continuous at zero in the usual norm induced Euclidean topology. A function is continuous if the pre-images of open sets are open. However, whereas $U := \mathbb{R}-\{1\}$ is an open set, $f^{-1}(U)=\{x\in \mathbb{R}\, : \,f(x)\neq 1\}= (-\infty,0]$ is not.
If we now change the topology to the discrete topology, where all sets are open, then $x \mapsto f(x)$ is all of a sudden continuous. So it's less the set which changes, it's its morphisms, resp. their properties, as the mapping didn't change in my example. So whether to consider one or the other topology, depends on what goals one has. The inner structure of a topological space changes if we change the topology.

 

[FactChecker]

So what's the difference if we take, say, the trivial topology and a function from the set to itself and another topology and the same function from the set to itself? I thought what makes it different was the open sets in which the function will act on.

For a function that maps from a set to the same set, there can be two different topologies defined for that set as a domain versus the same set as a range. It's very easy for the function to be continuous under one set of domain/range topologies and discontinuous under another. That is why I asked for more information about the function, the domain, and the range in post #2. Without specifying more about the function and the topologies, anything is possible.