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I have to grade some homework in solving the above question (in the title) but I have a dilemma.

QUESTION:

Given de Morgan's law and the fact that the intersection of a finite number of open sets is open, prove the union of a finite number of closed sets is closed.

ANSWER A:

For a family of closed sets $C_i$, whose union is $C = \cup_i^k\,C_i$, pick an $x \in C^c$ s.t. by de Morgan's law, $x \in \cap_i^k\,C_i^c$. There exists an $\epsilon > 0$ s.t. $B(x,\epsilon) \subseteq C_i^c$ for each $i$. So that for each $i$, $B(x,\epsilon) \cap C_i = \varnothing$ and so $B(x,\epsilon) \subset C^c$ showing that $x$ is an interior point of $C^c$. The point $x$ was arbitrary and so $C^c$ is open and hence $C$ is closed.

ANSWER B:

Suppose $C_i$ are all closed sets and want to prove $\cup_i^k\,C_i$ is closed, by proving $\left( \cup_i^k\,C_i\right)^c$ is open. Know that $C_i^c$ are open, and from the 2nd fact given, know that $C_1^c \cap C_2^c \cap \cdots $ is open too. Therefore $ \left(\cup_i^k\,C_i\right)^c$ is open and so $\cup_i^k\,C_i$ is closed


So my problem is that I was naively expecting the students to use points and balls to prove this, but ANSWER B reflects a common approach used by the students, with no mention of interior points, accumulation points, balls, etc. The subject matter in the text introduced closed sets by using acc. pts. and open sets by using int. points. The text does not mention de Morgan's law, though the question does. There is also a theorem in the text that says, A subset $G$ of $\mathbb{R}^n$ is open iff the complement of $G$ is closed. but it is not proven - that is left as a question that is not assigned to the students.

So in my mind they cannot use the theorem, and no one has even cited it. But I feel a little unjustified in taking points away because of the comments like the one bolded in the second answer - even though it wasn't proven, i.e. it is related to the theorem.


So, even though these are handed back to the instructor/professor first, I would like to address this (apparent to me) dilemma by asking for advice on how to grade the second answer, say out of 6 points.

P.S. wish there was a tag "grading"....

P.P.S. I also found lot of sites, like Prove that a finite union of closed sets is also closed but my question is a little more about grading...

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Grading is completely subjective and changes from one person to another. –  Asaf Karagila Apr 6 at 22:12
1  
There are some great comments so far, and I've only been grading a little while, so I'd like to do it as well as possible. Though I am good at being consistent, I do not want to shortchange anyone. :) –  nate Apr 6 at 22:15
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I suggest asking such questions at Mathematics Educators –  user127096 Apr 6 at 23:09
    
aaahhhh... didn't think to look for others. thanks –  nate Apr 6 at 23:48

3 Answers 3

up vote 2 down vote accepted

Full credit in my opinion. The way the question is worded seems to beg for that type of response. That the complemement of an open set is a closed set is almost the most basic definition of closed there is. It is the one I think of.

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Well, I actually like the answer B more, because it proves the statement for an arbitrary topological space, not only for a metric space. Full credit, I believe.

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1  
Yes, although A could be fixed to omit mention of radius and just use neighborhoods. –  augurar Apr 6 at 22:07
    
Brilliant point! Although I agree with augurar, something Topology students need to realize is that as soon as they start mentioning balls, any thing they say is limited to the arena of a metrizable topology. IIn light of this, I'd say A is not a full proof unless augurar's suggestion is made. –  Doop Apr 6 at 22:11

Keep in mind that the statement "A set is closed iff its complement is open" is usually a definition not a theorem. I personally feel that the second answer is the better one. So yes, full credit as Doop said.

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