# Extending continuous functions from dense subsets to quasicompacts

$$\DeclareMathOperator\cl{cl}$$I am interested under what assumptions one can always extend continuously a function defined on a dense subset; the range of the function is compact but not necessarily Hausdorff.

That is, I am interested in generalisations of the following theorem [Engelking, General Topology] to non-Hausdorff compact spaces:

3.2.1. THEOREM. Let $$A$$ be a dense subspace of a topological space $$X$$ and $$f$$ a continuous mapping of $$A$$ to a compact space $$Y$$. The mapping $$f$$ has a continuous extension over $$X$$ if and only if for every pair $$B_1$$, $$B_2$$ of disjoint closed subsets of $$Y$$ the inverse images $$f^{-1}(B_1)$$ and $$f^{-1}(B_2)$$ have disjoint closures in the space $$X$$.

I am mostly interested in sufficient conditions.

For example, is the following sufficient?

(i) For each $$Z_1, Z_2\subset A$$, it holds $$\cl_X(Z_1) \cap \cl_X( Z_2) = \cl_X( \cl_A(Z_1)\cap \cl_A(Z_2))$$.

(ii) For each $$Z\subset X$$ closed, and each pair of closed subsets $$Z_1, Z_2\subset A$$ such that $$Z\cap A=Z_1\cup Z_2$$, there are $$Z'_1, Z'_2 \subset X$$ closed such that $$Z=Z'_1\cup Z'_2$$, and $$Z_1=Z'_1\cap A$$, and $$Z_2=Z'_2\cap A$$, and $$Z'_1\cap Z'_2=\cl_X(Z_1\cap Z_2)$$.

(iii) $$A$$ is an open dense subset of $$X$$.

## 1 Answer

$$\DeclareMathOperator\cl{cl}$$Curiously enough, I was looking for a similar result, and I found Blair - Extensions of continuous functions from dense subspaces (and its errata) to be very helpful. Essentially, you define the Lebesgue sets of $$f\colon A\subset X\rightarrow Y$$ to be $$L_{a}(f)=\{x\in A \colon \;f(x)\leq a\}$$ and $$L^{a}(f)=\{x\in A \colon \;f(x)\geq a\}$$, and then you prove that there exists a continuous extension of $$f$$ from the dense subspace $$A$$ to $$X$$ if and only if $$a and $$\bigcap_{n=0}^{\infty}\,\cl_{X}(L_{-n}(f)\,\cup\,L^{n}(f))\,=\,\emptyset\,,$$ where $$\cl_{X}$$ denotes the closure in $$X$$. Note that both $$X$$ and $$Y$$ are generic topological spaces (no compactness is needed). I do not remember the proof, but the paper has it done in full detail.