I have a set of points $S = \{x_i\}_{i = 1}^m, x_i \in \mathbb{R}^n \forall i$. Now, I have a set function $f$ which operates as follows: $$f(S) = GX^T$$ where $G \in \{0,1\}^{m\times m}$ and $X = [x_1 x_2 \ldots x_m]$.

Is there a specific name for such $f$'s?

  • 2
    $\begingroup$ Set-valued function? $\endgroup$
    – Siminore
    Jul 22, 2013 at 7:51
  • $\begingroup$ The set-valued functions that I have come across don't seem to have set-valued input to the functions. I mean, the kind of set-valued functions studied seem to be $f:X\rightarrow 2^Y$ but not $f:2^X\rightarrow 2^Y$. $\endgroup$ Jul 22, 2013 at 8:05

1 Answer 1


The function $f$ does not return a set of points: it returns an $m\times 1$ matrix of real $n$-tuples. And the input is not a set $S=\{x_k:k=1,\dots,m\}\subseteq\Bbb R^n$: you’re using the indexing of the $x_k$ in an essential way, so the input to $f$ is actually an $m$-tuple of vectors in $\Bbb R^n$. Specifically,

$$f:\left(\Bbb R^n\right)^m\to\left(\Bbb R^n\right)^{m\times 1}:\langle x_1,\dots,x_m\rangle\mapsto G\begin{bmatrix}x_1\\x_2\\\vdots\\x_m\end{bmatrix}\;.$$

It’s simply a linear transformation between two finite-dimensional real vector spaces.

  • $\begingroup$ The reason I used the word set is because the size m of S is not necessarily fixed. $\endgroup$ Jul 22, 2013 at 8:13
  • $\begingroup$ Actually, hmm scratch that, possibly m can be fixed. I will have to think about it. Thanks! I will upvote for the moment, will accept once I get some clarity on the situation myself. $\endgroup$ Jul 22, 2013 at 8:14
  • $\begingroup$ @TenaliRaman: You’re welcome! $\endgroup$ Jul 22, 2013 at 8:17

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