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Nov
24
comment Help to find X (logic)
@Ofir: $X$ is the set consisting of the union of the orange and blue regions, but there is more than one way to describe that set in terms of $A,B$, and $C$. The most straightforward is $(C\setminus A)\cup B$. Another is $C\setminus(A\setminus B)$. Both of these are easy to see from the picture. There are others as well, but they’re more complicated.
Nov
24
answered Why $R=\{(a,b)\mid a=b \mbox{ or } a=-b\}$ is not anti symmetric?
Nov
24
answered Real Analysis: Continuity example
Nov
24
comment Help to find X (logic)
@user2976686: $X$ is the union of the orange and the blue. The orange is easy: it’s just $B$. What’s the blue? (It’s actually there in the diagram.)
Nov
24
comment Prove that (a) n < $k^+$ iff n ≤ k and (b) < is a transitive relation
@Stumpedphilosopher: You’re welcome. I’ve corrected the first paragraph so that it matches what you wrote, with the parentheses. No, $\varnothing\cup\{\varnothing\}=\{\varnothing\}$: the members of $\varnothing\cup\{\varnothing\}$ are the things that are in $\varnothing$ together with the things that are in $\{\varnothing\}$. There’s nothing in $\varnothing$, and the one thing in $\{\varnothing\}$ is $\varnothing$, so the one thing in $\varnothing\cup\{\varnothing\}$ is $\varnothing$; i.e., $\varnothing\cup\{\varnothing\}=\{\varnothing\}$.
Nov
24
revised Prove that (a) n < $k^+$ iff n ≤ k and (b) < is a transitive relation
Revised first paragraph, since I’d misread the question.
Nov
24
comment Help to find X (logic)
@user2976686: No, your answer should have the form $$X=\text{something involving }A,B,\text{ and }C\;.$$ (And in any case $A\cap C'=\varnothing$, since $A\subseteq C$, and $A\cup B'$ is the universal set, since $B\subseteq A$.)
Nov
24
comment Does the following question regarding functions and relations make sense?
@Dunno: You’re welcome.
Nov
24
comment Subnet vs. Subsequence
@user110894: You’re very welcome!
Nov
24
comment Prove the following $A\oplus B = \neg((A\cap B) \cup \neg(A\cup B))$
@Ofir: $X\setminus(X\cap Y)=X\cap\neg(X\cap Y)=X\cap(\neg X\cup\neg Y)=(X\cap\neg X)\cup(X\cap\neg Y)=\varnothing\cup(X\setminus Y)=X\setminus Y$. See if you can do the other one yourself: start with $(X\cup Y)\setminus Z=(X\cup Y)\cap\neg Z$ and use a distributive law.
Nov
24
answered Help to find X (logic)
Nov
24
answered Isometry of a metric space with proper subset
Nov
24
answered Does the following question regarding functions and relations make sense?
Nov
24
answered Prove that (a) n < $k^+$ iff n ≤ k and (b) < is a transitive relation
Nov
24
comment Proving Equivalence of DFA and NFA
@techno: There is no mistake. $M'$ is a DFA, and $Q'=2^Q$ is its state set, where $Q$ is the state set of the original NFA. The individual states of the DFA $M'$ are the sets of states of the NFA $M$.
Nov
23
answered Help with natural number problems
Nov
23
answered How to multiply permutations
Nov
23
answered Interval of convergence homework
Nov
23
answered Induction On $a_{n+1}$ Sequence
Nov
23
comment Induction On $a_{n+1}$ Sequence
Please verify that I correctly interpreted everything.