# Thread: How to prove the following containment relation about interiors

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

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

(A $\displaystyle \cup$ B)º $\displaystyle \supseteq$ Aº $\displaystyle \cup$ Bº

2. Here's one way to go about it. Pick an arbitrary $\displaystyle x\in\text{int}(A)\cup\text{int}(B).$ Then, WLOG, $\displaystyle x\in\text{int}(A).$ This implies that there is an $\displaystyle \varepsilon>0$ such that an open ball centered at $\displaystyle x$ of radius $\displaystyle \varepsilon$ is contained in $\displaystyle \text{int}(A).$ Now show that both $\displaystyle x\in A\cup B,$ and the ball is also contained in $\displaystyle A\cup B.$ That should do it, right?

3. Originally Posted by Ackbeet
Here's one way to go about it. Pick an arbitrary $\displaystyle x\in\text{int}(A)\cup\text{int}(B).$ Then, WLOG, $\displaystyle x\in\text{int}(A).$ This implies that there is an $\displaystyle \varepsilon>0$ such that an open ball centered at $\displaystyle x$ of radius $\displaystyle \varepsilon$ is contained in $\displaystyle \text{int}(A).$ Now show that both $\displaystyle x\in A\cup B,$ and the ball is also contained in $\displaystyle A\cup B.$ That should do it, right?
I'm confused on the part about the open ball. I don't think i've learned that yet.

4. 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 $\displaystyle 2\varepsilon$ centered at $\displaystyle x.$ In 2 dimensions, it'd be a circle of radius $\displaystyle \varepsilon$ centered at $\displaystyle x$ (using the usual Euclidean length). In 3D, it'd be a sphere of radius $\displaystyle \varepsilon$ centered at the vector $\displaystyle x.$ And so on. Does that make sense? It's a small open set containing $\displaystyle x.$ That's the critical part.

5. Originally Posted by Ackbeet
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 $\displaystyle 2\varepsilon$ centered at $\displaystyle x.$ In 2 dimensions, it'd be a circle of radius $\displaystyle \varepsilon$ centered at $\displaystyle x$ (using the usual Euclidean length). In 3D, it'd be a sphere of radius $\displaystyle \varepsilon$ centered at the vector $\displaystyle x.$ And so on. Does that make sense? It's a small open set containing $\displaystyle x.$ That's the critical part.

Ohhh, Yes! thank you

6. You're welcome! Were you able to finish the proof?

7. Originally Posted by Ackbeet
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 $\displaystyle 2\varepsilon$ centered at $\displaystyle x.$ In 2 dimensions, it'd be a circle of radius $\displaystyle \varepsilon$ centered at $\displaystyle x$ (using the usual Euclidean length). In 3D, it'd be a sphere of radius $\displaystyle \varepsilon$ centered at the vector $\displaystyle x.$ And so on. Does that make sense? It's a small open set containing $\displaystyle x.$ That's the critical part.
So wait can you write what you had described like this: x$\displaystyle \in$ (p-$\displaystyle \epsilon$,p+$\displaystyle \epsilon$) ? Sorry, I'm just trying to figure out a nice way to write out this proof.

8. 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.

9. If you're doing real analysis on the real line, you could write your neighborhood like this: $\displaystyle (x-\varepsilon,x+\varepsilon).$ You want to show that

$\displaystyle x\in A\cup B,$ and that

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

Right?