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Here's the problem:

Suppose that $S\subset \mathbb R^m$ is a convex set and $T\subset \mathbb R^n$ is a convex set. Show that the set $$S \times T = \{ (x_1 ,...,x_{m+n}\in \mathbb R^{m+n}):(x_1 ,...,x_m)\in S;(x_{m+1},...,x_{m+n})\in T\}$$ is a convex subset of $\mathbb R^{m+n}$.

So, I'm not very experienced with this, so bear with me. I know how to set up the argument, but not sure how to make the next step. I made two vectors: $$\vec x = (x_1,...x_m,x_{m+1},...,x_{m+n})$$ $$\vec y = (y_1,...y_m,y_{m+1},...,y_{m+n})$$

Now, I believe I need to show that $$ (\lambda x_1+(1-\lambda)y_1,...,\lambda x_m +(1-\lambda )y_m, \lambda x_{m+1} + (1-\lambda)y_{m+1},..., \lambda x_{m+n} + (1-\lambda)y_{m+n})\in S \times T $$

Here's where I'm stuck. I tried doing some factoring but nothing popped out to me. (Of course, it's also possible that I didn't approach this properly in the first place.) Any help would be appreciated. Thanks!

thisisourconcerndude
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1 Answers1

14

Let $(x_1,y_1), (x_2,y_2)\in S\times T$ and $t\in [0,1]$. Then $$(1 - t)(x_1,y_1) + t(x_2,y_2) = ((1 - t)x_1 + tx_2, (1 - t)y_1 + ty_2).$$

Since $S$ is convex and $x_1,x_2\in S$, so is $(1 - t)x_1 + tx_2\in S$; since $T$ is convex and $y_1,y_2\in T$, so is $(1 - t)y_1 + ty_2\in T$. Therefore $$((1 - t)x_1 + tx_2, (1 - t)y_1 + ty_2)\in S\times T,$$ which shows that $(1 - t)(x_1,y_1) + t(x_2,y_2)\in S\times T$.

kobe
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  • Could the same approach be taken to prove the product of two star-shaped sets is a star-shaped set, too? If the ordered pair of the vantage points $(p_0,p_1)$ is taken as $(x_1,y_1)$? – Matcha Latte Jun 23 '22 at 10:33
  • @Invisible yes, you are correct. – kobe Jun 23 '22 at 13:45