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QN1 by logical argument verify that {a} is not open for any real number "a".

i guess the set is not open since there is no open interval about "a" instead the set is said to be closed since any of its complement must be open. am i correct?


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Do not write using only capitals. It is considered shouting and thus impolite. As to your questions: 1. what is your definition of open? 2. Image under what map? – Glen Wheeler Apr 1 '11 at 14:36
@Glen: Doesn't matter which map... the question is to show that, in general, one does not necessarily have $f(A\cap B) = f(A)\cap f(B)$. – Arturo Magidin Apr 1 '11 at 14:57
@Arturo: really? It's certainly not true for all maps ($f(x) = 0$ for example). Doesn't it seem a bit odd? I can choose whatever map I like, any two sets (four sets) I like, any topology, any anything, and then just verify that with this choice $f(A\cap B) \ne f(A) \cap f(B)$? – Glen Wheeler Apr 1 '11 at 18:32
@Glen: The problem is badly stated, but it is a standard problem: show that, in general, it is not true that $f(A\cap B)=f(A)\cap f(B)$; that is, there exist sets $X$ and $Y$, a function $f\colon X\to Y$, and subsets $A$ and $B$ of $X$ for which $f(A\cap B)\neq f(A)\cap f(B)$. (This in contrast with showing that for all sets $X$ and $Y$, all functions $f\colon X\to B$, and all subsets $A$ and $B$ of $X$, $f(A\cup B)=f(A)\cup f(B)$, for example, or that equality holds for inverse image of the intersection). So, yes: the point is to come up with a single example where it doesn't work. – Arturo Magidin Apr 1 '11 at 18:35
@Glen: P.S., no topology needed; this is about the basic set-theoretic functions of direct and inverse image associated to any set-theoretic function. The inverse image map is well-behaved, but the direct image map is not. – Arturo Magidin Apr 1 '11 at 18:52

If you are working with the usual topology on the reals, then the complement of {a} is the union of two open intervals and so open. If {a} were open, its complement would need to be closed, which it is not (a is a limit point of the complement, but not included in the complement). So, your argument for question 1 seems OK.

For question 2, you could always take the cheap way out by letting S and T be nonempty disjoint subsets of the reals and f a constant function. The image of the intersection of S and T is then empty, as is the image of f on the intersection. The intersections of the images is whatever the constant value of f is. Another possibility is to pick a function like $f(x) = x^{2}$ and take S=[-1,0] and T=[0,1].

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Hint for Q2:

Must the image of a point in the intersection of two sets be in the intersection of the images of the two sets?

Must a point in the intersection of the images of two sets be the image of a point in the intersection of the two sets?

If the answers to these questions are different, then can you construct a counter-example?

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