Suppose $f:[0,1]\times[0,1]\mapsto X$ is a continuous function. Show that $[0,1]\times[0,1]$ can partitioned into rectangles s.t $f(R_i)\subseteq U_k$

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 $$U_i$$.

• Oh right so this definitely doesnt work. – AColoredReptile Mar 21 at 23:45
• Partition R = [0,1]×[0.1] into {R} . Thus f(R) subset open X and {X} is an open cover of X. – William Elliot Mar 22 at 1:26
• Do you know the Lebesgue Number Lemma? – Lee Mosher Mar 22 at 2:02
• @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. – 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 $$f(R) \subset U_i$$.