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Math Help - How to prove the following containment relation about interiors

  1. #1
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    How to prove the following containment relation about interiors

    How would I prove the following containment relation about the interior of a union:

    (A \cup B) \supseteq A \cup B
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  2. #2
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    Here's one way to go about it. Pick an arbitrary x\in\text{int}(A)\cup\text{int}(B). Then, WLOG, x\in\text{int}(A). This implies that there is an \varepsilon>0 such that an open ball centered at x of radius \varepsilon is contained in \text{int}(A). Now show that both x\in A\cup B, and the ball is also contained in A\cup B. That should do it, right?
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    Quote Originally Posted by Ackbeet View Post
    Here's one way to go about it. Pick an arbitrary x\in\text{int}(A)\cup\text{int}(B). Then, WLOG, x\in\text{int}(A). This implies that there is an \varepsilon>0 such that an open ball centered at x of radius \varepsilon is contained in \text{int}(A). Now show that both x\in A\cup B, and the ball is also contained in A\cup B. That should do it, right?
    I'm confused on the part about the open ball. I don't think i've learned that yet.
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  4. #4
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    You can call it a neighborhood, or a sphere, or whatever you want. On the real line, it'd be an open interval of length 2\varepsilon centered at x. In 2 dimensions, it'd be a circle of radius \varepsilon centered at x (using the usual Euclidean length). In 3D, it'd be a sphere of radius \varepsilon centered at the vector x. And so on. Does that make sense? It's a small open set containing x. That's the critical part.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    You can call it a neighborhood, or a sphere, or whatever you want. On the real line, it'd be an open interval of length 2\varepsilon centered at x. In 2 dimensions, it'd be a circle of radius \varepsilon centered at x (using the usual Euclidean length). In 3D, it'd be a sphere of radius \varepsilon centered at the vector x. And so on. Does that make sense? It's a small open set containing x. That's the critical part.

    Ohhh, Yes! thank you
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  6. #6
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    You're welcome! Were you able to finish the proof?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    You can call it a neighborhood, or a sphere, or whatever you want. On the real line, it'd be an open interval of length 2\varepsilon centered at x. In 2 dimensions, it'd be a circle of radius \varepsilon centered at x (using the usual Euclidean length). In 3D, it'd be a sphere of radius \varepsilon centered at the vector x. And so on. Does that make sense? It's a small open set containing x. That's the critical part.
    So wait can you write what you had described like this: x \in (p- \epsilon,p+ \epsilon) ? Sorry, I'm just trying to figure out a nice way to write out this proof.
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  8. #8
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    In place of ball just say there is an open set that contains x and is a subset of A. That works in any topological space for interior points.
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  9. #9
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    If you're doing real analysis on the real line, you could write your neighborhood like this: (x-\varepsilon,x+\varepsilon). You want to show that

    x\in A\cup B, and that

    (x-\varepsilon,x+\varepsilon)\subseteq A\cup B.

    Right?
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