Suppose $f:[0,1]\times[0,1]\mapsto X$, is a continuous function where $X$ compact and connected subset of $\mathbb{R}^n$. Show that $[0,1]\times[0,1]$ can partitioned into rectangles $R_i$ such that $f(R_i)\subseteq U_i$ where $U_i\in C$ a cover for $X$.

My attempt at a proof:

Choose a cover for $X$, then it has a finite subcover.

Suppose not, then for any rectangles with side lengths $x\in \mathbb{N}$, $f(R_i)\not\subseteq U_i$ for any $U_i\in C$. I believe I want to take rectangles of side lengths, $\frac{1}{n}$ and then make a sequence of balls of radius $\frac{1}{m}$, for $m\in \mathbb{N}$ and then use sequential compactness to get the image can be contained in a single $U_i$.

  • $\begingroup$ Oh right so this definitely doesnt work. $\endgroup$ – AColoredReptile Mar 21 at 23:45
  • $\begingroup$ Partition R = [0,1]×[0.1] into {R} . Thus f(R) subset open X and {X} is an open cover of X. $\endgroup$ – William Elliot Mar 22 at 1:26
  • $\begingroup$ Do you know the Lebesgue Number Lemma? $\endgroup$ – Lee Mosher Mar 22 at 2:02
  • $\begingroup$ @Lee Mosher Yes, the proof I know of it is similar to the one I am trying to do, but I'm not sure how to make it work for the image of a function. $\endgroup$ – AColoredReptile Mar 22 at 2:34

The easiest thing is to simply apply the Lebesgue Number Lemma to the open covering of $X$ that you get by pulling back the given open covering of $Y$, namely $${\cal V} = \{f^{-1}(U_i) \mid U_i \in C\} $$ If you apply the lemma you get a Lebesgue number $\lambda>0$. Then subdivide $[0,1] \times [0,1]$ into rectangles whose diagonal length is $<\lambda$. Each little rectangle $R$ will be contained in some $f^{-1}(U_i) \in \cal{V}$, and so $f(R) \subset U_i$.


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