# 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 $\cup$ B)º $\supseteq$ $\cup$

2. 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?

3. Originally Posted by Ackbeet
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.

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 $2\varepsilon$ 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.

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 $2\varepsilon$ 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

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 $2\varepsilon$ 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.

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: $(x-\varepsilon,x+\varepsilon).$ You want to show that

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

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

Right?