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I am having some doubts about enumerating elements of this set:

$$\overline{X} = \{10-x \ \ |\ \ x \in \varnothing\}$$

Can you enumerate what elements are in there? I know this is a really elementary question, I'm sorry. According to me it should be $\overline{X} = \{10\}$ but I'm not really sure. Care to explain? Thanks!

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  • $\begingroup$ there is a problem with the formula $x \in \emptyset$. By definition (roughly) the empty set is the set of zero cardinality with no elements. So this formula doesn't make any sense $\endgroup$ – user321268 Jun 17 '16 at 16:02
  • $\begingroup$ In an exercise I found that someone wrote down that $\overline{X} = \{0\}$ but this looks non-sense to me. $\endgroup$ – user1365914 Jun 17 '16 at 16:03
  • $\begingroup$ However, since the empty set has no elements, except the "$x \in \emptyset$" part, someone can claim that, it's true that $X= \{ 10-x | \forall x \in \emptyset\} = \{10\}$ $\endgroup$ – user321268 Jun 17 '16 at 16:07
  • $\begingroup$ @mayer_vietoris: The formula $x \in \varnothing$ makes sense, it's just false for all values of $x$. Thus any set of the form $\{ \text{expression} \mid x \in \varnothing \}$ is empty, as in PHPirate's answer. $\endgroup$ – Clive Newstead Jun 17 '16 at 16:14
  • $\begingroup$ @mayer_vietoris The formula makes perfect sense to me. The set of all 10 - x where x does not exist is the empty set. That a formula might have no instances doesn't mean it doesn't make sense. Take Y = {27p + 5a| a in N; 3|a; a < 643; p in N; p prime; 2|p; p > 17} This formula makes perfect sense even though there is no possible p and therefore no possible instances. $\endgroup$ – fleablood Jun 17 '16 at 16:17
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$\overline X \ne \{10\}$ because $0 \not \in \emptyset$. You are sort of thinking of this as "start with 10; now subtract x for each x in emptyset and keep track of the answers". This isn't actually right as there is no "start with 10" part. It's actually more, "start with nothing (not zero, nothing, i.e. start with the emptyset) and then for each x in emptyset take 10 - x and keep track of your results". You'll end up with nothing (you won't ever get started).

To be obtuse but a bit more formal then PHPirate's excellent intuitive answer...

If $y \in \overline X$ then $y = 10 - x$ for some $x \in \emptyset$. At this point it should be clear that this is impossible. so it is impossible that $y \in \overline X$ so $\overline X = \emptyset$.

To be really obtuse I could say: as $y = 10 -x; x \in \emptyset$ is impossible, so $y \in \emptyset$. Therefore $\overline X \subseteq \emptyset$. But $\emptyset \subseteq \overline Y$ ($\emptyset \subseteq$ any set) so $\overline X = \emptyset$.

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When I read it out loud, I'm saying "All elements $10-x$ with $x$ not existing". So since there are no $x$'s, there are also no elements $10-x$ I think. So $\overline{X} = \emptyset.$ Your thoughts seem to be about $x \in \{0\}$.

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Reading the first part of the definition I see $10-x$ so I suppose that $x$ is an element of some set of numbers. So, in the second part I 'm forced to think at $\varnothing$ as the void subset of such set of numbers that, by definitions, contains no numbers. So I'm tempted to agree with the OP that $\bar X=\{10\}$, but the problem is that a binary operation (as $10-x$) require two numbers as arguments and in this case $x$ is not a number, so I think that $\bar X$ has not well defined elements and it is void.

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The notation $\{ f(x) \mid \varphi(x) \}$ where $f$ is a function, and $\varphi$ is a formula, is really shorthand for $$\{ y \mid y = f(x)\ \wedge\ \varphi(x) \}$$ That is, $\{ f(x) \mid \varphi(x) \}$ is the set of things of the form $f(x)$, where the formula $\varphi(x)$ is true.

So your set $\overline X$ is equal to $$\{ y \mid y=10-x\ \wedge\ x \in \varnothing \}$$ Now the formula $x \in \varnothing$ is always false, so this set has no elements. That is, $\overline X = \varnothing$.

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