How do I show that the open rectangle is convex?

Question

I have no idea how to solve this. It's easy to show that the open ball or the open cube is convex, but how do we show that the open rectangle is? (The same holds for the closed rectangle).

Thanks

Answer 1

I assume that you are working in a vector space. Let's prove that if $U_1 \subseteq V_1$ and $U_2 \subseteq V_2$ are convex, where $V_i$ are vector spaces, then $U_1 \times U_2 \subseteq V_1 \times V_2$ is also convex. This is sufficient, since every interval in the real line is convex, and rectangles are finite products of intervals.

Let $(x,y),(x,y') \in U_1 \times U_2$. Let $t \in [0,1]$ be arbitrary. Since $x,x' \in U_1$ and $U_1$ is convex, we have $tx+(1-t)x' \in U_1$. Since $y,y' \in U_2$ and $U_2$ is convex, we have $ty+(1-t)y' \in U_2$. Hence $$t(x,y)+(1-t)(x',y') = (tx+ (1-t)x',ty+(1-t)y') \in U_1 \times U_2.$$

Answer 2

I am reading "Analysis on Manifolds" by James R. Munkres.
This exercise is the same exercise as Exercise 4(b) on p.40 in this book.

Let $t\in [0,1].$

Let $\mathbf{c},\mathbf{d}\in (a_1,b_1)\times\dots\times(a_n,b_n).$
Then, $$a_i<\min\{c_i,d_i\}\leq (1-t)c_i+td_i\leq\max\{c_i,d_i\}<b_i.$$ So, $\mathbf{b}+(1-t)\mathbf{c}\in (a_1,b_1)\times\dots\times(a_n,b_n).$

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Let $\mathbf{c},\mathbf{d}\in[a_1,b_1]\times\dots\times[a_n,b_n].$
Then, $$a_i\leq\min\{c_i,d_i\}\leq (1-t)c_i+td_i\leq\max\{c_i,d_i\}\leq b_i.$$
So, $\mathbf{b}+(1-t)\mathbf{c}\in [a_1,b_1]\times\dots\times [a_n,b_n].$