convergence of compact sets

Question

Let $X$ and $Y$ be compact sets (both subsets of the real line). Assume $Y$ has a non-empty interior. Consider the continuous function $f:X \rightarrow Y$. For any given $y$ in the interior of Y, and for $h>0$ small enough so that $y+h \in Y$. I want to know whether it is true that $f^{-1}([y,y+h]) \rightarrow f^{-1}(y)$ when $h \rightarrow 0$

I know that $f^{-1}([y,y+h])$ is a closed set for each $h$. My main problem is that we are talking about convergence between two sets. How do you define convergence between two sets? More informally, my question is does the sequence of set converges to the set composed of the inverse of image of $f(y)$

help?

Answer 1

Let $A(h)=f^{-1}([y,y+h])$. Clearly, $A(h)\subset A(k)$ if $h<k$ (Why?). For such a family of sets, there are two limits you might consider: $A_{\infty}=\cup_{h>0} A(h)$ or $A_{0}=\cap_{h>0} A(h)$. The one you have in mind would by the intersection, since this gives you the set that remains upon taking $h$ to zero. If there is an $x\in A_0$, what must be true of $x\in \cap_{h>0} A(h)$? Once you write down what condition such an $x$ must satisfy, you will see that $f(x)=y$, i.e., $x\in f^{-1}(\{y\})$.

Note that all of that doesn't require $f$ to be continuous, $X$ or $Y$ to be compact. Those assumptions are useful for establishing the fact that $A_0$ is non-empty.

(By the way, for a compact set $Y$ and an element $y\in Y$ there does not need to exist an $h>0$ (small or otherwise) such that $y+h\in Y$. Perhaps you have left out some assumptions on $Y$ and $y$?)