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Prove that the following is true:

$$int(A)∩int(B)=int(A∩B)$$

I'm in general really bad at writing proofs. I do not know where to start even when looking at the definitions given in the textbook. I am well aware of the definition of the interior of sets, but I can't apply it in a meaningful way. My attempt at writing that proof is laughable since I'm doing the proof backwards.

We know that $int(A)\subset A$, thus we have:

$$int(A∩B)=int(A)∩int(B)\\ int(A\cap B)\subseteq A \cap B$$

The latter is per definition the interior of the set $A \cap B$.

Q.E.D.

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    Is this happening in a general topological space or on $\mathbb{R}^n$? – 2017-01-18
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    The two sets are arbitrary – 2017-01-18

2 Answers 2

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I can't make any sense of your proof - it seems like you use what you are trying to prove in the first line.

Here are some hints:

  1. You will have to use the definition of interior. Since this looks like an introductory proofs class, I assume that you are using the definition: $x$ is in the interior of $A$ if there exists an open ball centered at $x$ that is contained in $A$.

  2. The easiest way to show that two sets are equal is usually to show that the first is contained in the second, and then show that the second is contained in the first. So, you can first try to show $int(A \cap B) \subseteq int (A) \cap int(B)$, and then show the opposite inclusion.

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    This is a very good point: the fact that the OP's attempted proof makes _no_ (explicit) use of the definition of interior should immediately be a red flag. – 2017-01-18
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    I've managed to write the proof on paper now. I got confused on the definition of an open ball. More precisely the point in which the ball is centered around. Now it's all fine and dandy! – 2017-01-18
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Doing a proof backwards does not have to be a bad idea. It's often a useful tool to use when trying to figure out how to prove it, only to work it out backwards later. In your case however, you use what you're trying to prove as a given.

Try proving that $$int(A∩B)\subseteq int(A)∩int(B)$$ and that $$int(A)∩int(B)\subseteq int(A∩B)$$