Criteria for extension to cts fn between metric spaces?

[some_dude]

Given metric spaces

$$(X, d_X), (Y, d_Y)$$,

and subsets

$$\{ x_1, ..., x_n \}, \{ y_1, ..., y_n \}$$

of X and Y respectively, if I define a function that send $$x_i$$ to $$y_i$$, what are the minimal restrictions we need on X and Y in order for there to exist an extension of that map to a continuous function defined on all of X?

Certainly you'd need some relationship between the connected components of both. So for simplicity, also assume each of the metric spaces are (path) connected. Also assume X and Y are compact.

If you make any assumptions on Y that will allow you to apply Uryshon's lemma or Tietze extension theorem, then it is not very interesting. I'm curious about the minimal set of restrictions on Y needed to find this extension. (Perhaps I'm missing some elementary theorem that would give my answer?).

 

[Eynstone]

Given metric spaces

$$(X, d_X), (Y, d_Y)$$,

and subsets

$$\{ x_1, ..., x_n \}, \{ y_1, ..., y_n \}$$

of X and Y respectively, if I define a function that send $$x_i$$ to $$y_i$$, what are the minimal restrictions we need on X and Y in order for there to exist an extension of that map to a continuous function defined on all of X?

Certainly you'd need some relationship between the connected components of both. So for simplicity, also assume each of the metric spaces are (path) connected. Also assume X and Y are compact.

If you make any assumptions on Y that will allow you to apply Uryshon's lemma or Tietze extension theorem, then it is not very interesting. I'm curious about the minimal set of restrictions on Y needed to find this extension. (Perhaps I'm missing some elementary theorem that would give my answer?).

If xi,xj lie in the same connected component, so should yi,yj : we can relax the condition of path-connectedness with this restriction.
Also, I think we need the condition of second countability on Y to construct such a function. I wonder if completeness of Y and X is necessary.

 

[some_dude]

If xi,xj lie in the same connected component, so should yi,yj : we can relax the condition of path-connectedness with this restriction.
Also, I think we need the condition of second countability on Y to construct such a function. I wonder if completeness of Y and X is necessary.

Thanks for that.

I'll point out relaxing path-connectedness would fail if you didn't assume Y were compact. E.g., Y = "Topologist's Sine Curve" (see wikipedia), and X = [0, 1]. Then if y_1 = (0,0), y_2 = some other arb pt in Y, and x_1 = 0, x_2 = 1. You'd be unable to extend that to a continuous function, as the compact interval could not be mapped onto the non-compact connected set containing y_1 and y_2.

I asked a mathematician about my original question, and, well, there response would indicate it's unlikely there's a cut and dried answer to be found.

 

[Eynstone]

I asked a mathematician about my original question, and, well, there response would indicate it's unlikely there's a cut and dried answer to be found.

In a way, that's right. If X has a property preserved under continuous maps, Y must have the same. The question can be answered only if one specifies what X & Y are.
I found the question interesting because if extended to countably many points, it would yield a telling result for separable metric spaces.

 

[blue_raver22]

I would approach this question by first examining the properties of continuous functions and how they relate to metric spaces. In order for a function to be continuous, it must preserve the topology of the metric space. This means that for any open set in the domain, the preimage of that set under the function must also be open.

In this case, we are looking for an extension of a function from a subset of X to all of X. This means that the preimage of any open set in X must also be open in the subset. Therefore, in order for an extension to exist, we need the subset to be dense in X. This ensures that the preimage of any open set in X will also be open in the subset, and thus the function will preserve the topology of X.

Additionally, we need the subset to be compact. This is because for a function to be continuous, it must also be uniformly continuous. This means that for any given epsilon, there exists a delta such that the distance between any two points in the domain is less than delta will result in the distance between the images of those points being less than epsilon. In order for this to hold for all points in the subset, it must be compact.

Finally, in order for the extension to be unique, we need the subset to be closed. This ensures that the function is well-defined and does not have any gaps or overlaps.

In summary, the minimal restrictions on X and Y for an extension to exist are that X must be a compact metric space and the subset must be dense, compact, and closed. These conditions ensure that the function will preserve the topology of X and be uniformly continuous, allowing for a unique and continuous extension to be defined on all of X.