Prove, that the only sets in $R^1$ which are both closed and open are $R^1$ and empty set(proof-check)
Consider instead $mathbbQ$ (with the relative topology from $mathbbR$). The set $S=xinmathbbQ:x^22 $$ is open as well.It follows that your attempt at a proof is wrong: if it worked, it would also apply to this situation, but it does not . You must use the specific property that makes the result true in $mathbbR$, namely that every upper bounded set has a supremum.
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Intersection of a non-empty set of natural numbers (set-theoretic definition) is a natural number?
Let $x$ be $bigcap E$. All elements of $E$ are natural numbers and thus all elements of $E$ are sets of natural numbers. So, $x$ is a set of natural numbers.As $E$ is not empty, there's a natural number $n_0 in E$. Clearly, $x subseteq n_0$.If $y in x$, then $y in n$ for all $n in E$. As all $n in E$ are transitive, we have $y subseteq n$ for all $n in E$, so $y subseteq bigcap E = x$, i.e. $x$ is transitive.Now you have that $x$ is a transitive set of natural numbers which is a subset of some natural number $n_0$. Does that suffice?
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Name of a set of the form x,y
A set with no more than two elementsIf you want to exclude the empty set then a non-empty set with no more than two elements
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Do the empty set AND the entire set really need to be open? [closed]
If one wants constant functions to always be continuous, then one must necessarily have the empty set and the whole space to be open.From a category theory perspective, it is the continuous functions that are the more fundamental building block of topology, than the open or closed sets. I believe that there is some equivalent way to axiomatise topology via continuous functions using the machinery of sheaves, which is in some ways more "natural" than the simple but somewhat arbitrary-looking axioms for open sets, but I am not an expert on these matters.
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Quantification over the empty set [duplicate]
Well if $forall y in emptyset : Q(y)$ were false, we would be able to find some $y in emptyset$ such that $Q(y)$ were false. However, there are no $y in emptyset$. So $forall y in emptyset : Q(y)$ should be true
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How can the empty set be open in the definition of the cofinite topology?
No it is not: Let $A=Xsubset X$, then $X backslash X = emptyset$ which is clearly finite, indeed $|emptyset|=0$, i.e. the cardinality of the empty set is zero. Hence $Xin T$
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Can definition be existence (in mathematics)?
Set theories need not postulate the a priori existence of any objects or structures. ZFC does, however, postulate the existence the empty set (it's zero) and a kind of successor function based on the empty set as a starting point. The resulting set could have infinitely many junk terms that need to selected out using the Separation (Subset) Axiom, leaving only the set of natural numbers, i. e. a subset that satisfies Peano's axioms. You could also simply postulate the existence of some Dedekind-infinite set outside of set theory -- not a huge leap of faith. (If such a set does not exist, the universe would be a very dull place indeed.) Then you can extract a subset from it that satisfies PA. Having shown the existence of a set that satisfies PA by one of these means, you would be quite justified in beginning your development of number theory and analysis by simply defining the natural numbers using PA
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What interpretations exists for the singleton of the singleton of the empty set $varnothing$ and its singleton $varnothing$?
You are taking the Kuratowski ordered pair definition too seriously. It does not have any intrinsic intuitive meaning, and in particular does not tell us anything deep about the sets it chooses to represent ordered pair. Its only role is as a technical detail in the proof ofIn the entire rest of the development of set theory (including the very formulation of several of the axioms), and of the development of the rest of mathematics as an application of set theory, we forget completely which formula this $phi(x,y,p)$ actually is and just write it as $langle x,y
angle=p$ instead.This involves taking care that we never depend on any properties of ordered pairs other than what the metatheorem guarantees -- in particular we never put ordered pairs and other things into the same set and expect to be able to tell them apart later, because a priori anything might be a pair. We could have proved the ordered-pair metatheorem for a different $phi$ -- for example one that represents the particular pair $langle omega_3,42,19
angle$ as $varnothing$ and everything else as Kuratowski pairs in reverse order -- and that should not make any difference at all for the rest of the development.So yes, it is true in a technical sense that $langle 0,0
angle $ and $1$ are the same set. But that is just a coincidence -- an implementation detail -- and we deliberately do not depend on such details when we do mathematics.It makes very good sense, as Asaf suggests, to view this as a type system. In that view, asking whether $A=B$ when we know $A$ as an ordered pair, and $B$ as a set with particular members, is a type error which produces an implementation-dependent (even if definite) truth value.