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[SOLVED] Interior of a set

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,104
The Astral plane
Here's the problem:
Let X be the plane with the usual topology while A = E = the x-axis. Then \(\displaystyle Int_A E = A\) while \(\displaystyle Int_X E = \emptyset\), so that the former cannot be obtained by intersecting the latter with A. It is always true, however, that \(\displaystyle Int_A E \supset A \cap Int_X E\)
I'm taking all of this to be that \(\displaystyle E \subseteq A \subseteq X\) and that X is simply \(\displaystyle \mathbb{R} ^2\) (with it's usual topology.)

I understand the final equation in the above. I haven't proven it in general but it easily satisfies the mental image that I have. (I probably should make sure to prove it, but this has little to do with the question I'm going to ask.)

So, the question I'm going to ask... Why is \(\displaystyle Int_X E = \varnothing\)? Every open interval in E can be found to be the intersection of an open set in X and E.

Note: My text defines
If X is a topological space and \(\displaystyle E \subset X\) then \(\displaystyle Int _X E = \cup \{G \in X | \text{G is open and } G \subset E \}\)
Thanks!

-Dan
 

Olinguito

Well-known member
Apr 22, 2018
251
Hi Dan.

Every open interval in E can be found to be the intersection of an open set in X and E.
That is true but is nothing to do with the definition of the interior of a set. $\mathrm{Int}_XE$ is the set of all open sets of $X$ contained in $E$; as all nonempty open sests of $X$ contain points outside the $x$-axis, there can be no nonempty open sets of $X$ contained in $E$.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,104
The Astral plane
Hi Dan.



That is true but is nothing to do with the definition of the interior of a set. $\mathrm{Int}_XE$ is the set of all open sets of $X$ contained in $E$; as all nonempty open sests of $X$ contain points outside the $x$-axis, there can be no nonempty open sets of $X$ contained in $E$.
Okay. Now that I think about it again I'm thinking I was trying to look at the converse.

Thanks!

-Dan
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Rather than "set of all open sets" better wording would be "union of all open sets". The interior of a set is a single set, not a collection of sets.