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The author writes $(X-C)\cap Y = Y-A,$ and, also, $A=Y \cap (X-U)$. I was wondering how is that something anyone writing an original proof of the theorem saw and if there is an analytic proof that the equalities holds true. Thanks for your help.

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3 Answers 3

up vote 2 down vote accepted

Is your question really "how does someone see that $(X - C) \cap Y = Y - A$ and $A = Y \cap (X - U)$"? Drawing a picture definitely helps; and in fact the book by Munkres already has two figures showing exactly that:

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Once you've seen this, a formal proof is not difficult anymore: either follow the computation by Frank in his answer, or just take an element in the set on the left and show it's in the set of the right and visa versa, as in Henno's answer.

I'm not sure what you mean by an analytical proof. If it's something with $\epsilon$'s and $\delta$'s as in analysis, then, no, such a proof doesn't exist, because you're working with general topological spaces here. If you mean some form of computation, then Frank's or Henno's answer give one.

Note, by the way, that the result of Theorem 17.2 and its proof are not very surprising. The topology on $Y$ is defined in such a way that the open sets in $Y$ are exactly the sets $Y \cap U$ with $U$ open in $X$. If you're wondering what the closed sets in $Y$ are, the first guess would be that they are the sets of the form $Y \cap C$ with $C$ closed in $X$ and that's exactly what this theorem states. The proof just translates between "$A$ is closed" and "the complement of $A$ is open" and uses the definition of open set in $Y$.

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Let me explain the first equality:

$Y \setminus A$

$= Y \setminus(C \cap Y)$

$= (Y\setminus C) \cup (Y \setminus Y)$

$= (Y \setminus C) \cup \emptyset$

$= Y\setminus C$

$= (Y \setminus C) \cap (X\setminus C)$

$= (Y \cup X) \setminus C $

$= X \setminus C$.

Equalities 1 and 6 are the result of De Morgan's Law. Equality 5 holds because $Y \setminus C \subseteq X \setminus C$. Equality 7 holds because $Y \subseteq X$ so that $Y \cup X = X$.

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He defined $A = C \cap Y$; To see $(X-C) \cap Y = Y - A$, we can compute (as Frank did) or reason: a point is in the left hand side if it is not in $C$ but it is in $Y$. This means not in $A$ (to be in $A$, one has to be both in $C$ and $Y$), so it is in $Y - A$. On the other hand, if a point is in $Y - A$, it is in $Y$ but not in $A$, and the only way that could happen is if it is not in $C$ (otherwise it would be in $A$ by definition). So it is not in $C$ (so in $X-C$) and in $Y$, so in $Y \cap (X-C)$.

Or one could draw a schematic picture, and see it that way (it's quite obvious then).

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