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In general, if B contains the origin of $\mathbb{E}$ then, the set erosion and dilation by $B$ become, respectively, anti-extensive and extensive; i.e., for all $X$: $ \begin{equation} X \ominus B \subseteq X \subseteq X \oplus B \end{equation}$

What I'm asking is, if B does not contain the origin of $\mathbb{E}$, is there such a B so that $X \oplus B \supseteq X \ominus B$ ISN'T always true?

Nyquist-er
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  • What has this to do with computer vision? More importantly, what do all the symbols mean? – FShrike Sep 25 '22 at 14:24
  • @FShrike Oldschool computer vision used to deal with lots of morphological filters (i.e dilation, erosion,opening and closing to name a few). The circle with a minus in it means erosion and the one with the plus in it means dilation. https://en.wikipedia.org/wiki/Mathematical_morphology – Nyquist-er Sep 25 '22 at 14:33
  • Hint: try $B = { b }$ for some $b \in \mathbb{E}$ different from the origin. – Adayah Sep 25 '22 at 14:33
  • Please do not add tags that are irrelevant to the question. – Asaf Karagila Sep 27 '22 at 01:04
  • @AsafKaragila Would you explain why you deem erosion and dilation (both being set operators) not relevant to set theory? – Nyquist-er Sep 27 '22 at 11:22
  • Have you looked at the tag description? The matrices in the answer contain numbers, why is this not a matrix theory? Or number theory? – Asaf Karagila Sep 27 '22 at 12:15
  • you can tackle this without numbers... please do not change my tag again so someone who happens to actually know about the SET operators dilation and erosion will see this. Have a good day! – Nyquist-er Sep 27 '22 at 12:24

1 Answers1

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The assumption of the erosion and dilation being, respectively, anti-extensive and extensive is only true if and only if the origin is contained in the structuring element.

For example, take a 5x5 zero-matrix with a 1 in the center then we have that dilation and erosion with structuring element,

$$ B = \left[ \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right] $$

produce the following matrix:

$$ \left[ \begin{array} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array} \right] $$

As you see $A \ominus B \not\subseteq A \not\subseteq A\oplus B$.

Edit: Wrong answer but you can solve your question by elemental set theory, if you see that $A \ominus B = \{z / (B_z) \subseteq A \}$ and $ A \oplus B = \{ z / (B_z)\cap A \subseteq A \} $ and take $z\in A \ominus B $ then:

$$ (B_z)\subseteq A \implies (B_z)\cap A \subseteq A \cap A = A$$

Then $z \in A\oplus B$ therefore $A \ominus B \subseteq A \oplus B$.

Cofinite
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