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Could I please have an example of closed sets with empty interior? Any topological space. Everything goes.

Remark: This is not homework. I'm in the middle of proving the space of $n$-degree polynomials in $(C[0,1], \lVert \cdot \rVert_\infty)$ is meager. Empty interior of this set (if I'm at all correct...) is the punchline and I just want to understand a bit more the nature of closed sets with empty interior.

Marios Gretsas
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Friedman
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  • Any finite-dimensional linear subspace of a Banach space is closed. Any closed linear proper subspace of a Banach space has empty interior. – DanielWainfleet Dec 27 '16 at 23:41

8 Answers8

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Here are a few intuitive examples

  • Every singleton set $\{p\}$
  • The circle $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = r^2\}$.
  • The line $y=mx + b$

The latter two can be generalized of course.

Edit: Here's a bit more intuition: the plane in $\mathbb{R^3}$ enter image description here

Take some point that doesn't lie in the plane. Do you see how you can find an open ball around it which doesn't intersect the plane? Think of the plane as being "thin", if you will. This tells you its complement is open, so the plane is closed. And by the "thin-ness", no point on the plane has an open ball around it contained in the plane.

So now we have a point, a line, and plane as being closed nowhere dense sets in $\mathbb{R}, \mathbb{R^2}, \mathbb{R^3}$ respectively. Can you try to generalize to find a closed set with empty interior in $\mathbb{R^n}$?

David Bowman
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Consider every finite set in the euclideian topology in $\mathbb{R}$

Also $\mathbb{Z}$ is closed because $\mathbb{R}$ \ $\mathbb{Z}=\bigcup_{n \in \mathbb{Z}}(n,n+1)$

Or the set $A= \{1/n | n \in \mathbb{N}\} \cup \{0\}$

Or the Cantor set which does not contain an interval but it is closed as an intersection of closed sets.

Or Every line and curve in the plane.

All these sets of course with the euclideian topology.

Also the finite sets in $\mathbb{N}$ with respect to the cofinite topology.

Marios Gretsas
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13

I'm surprised no one proposed this one : $\emptyset$.

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    As far as I am concerned, I am surprised no one proposed this one: Let $E$ be a normed vector space and $F\subseteq E,F\neq E$ be a finite-dimensional subvector space, then $F$ is closed and has empty interior. – C. Falcon Dec 27 '16 at 00:45
8

Consider in $\mathbb{R}^2$ the straight line $\{(x,x) \in \mathbb{R}^2 \; | \; x \in \mathbb{R}\}$. This set is closed since its complement is open, but it contains no ball in the Euclidean topology of $\mathbb{R}^2$.

Mnifldz
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8

Consider $X=\{a,b,c\}$ and $\tau=\{X,\varnothing,\{a\}\}$ and let $A=\{b,c\}$. Then $A$ is closed but $int(A)=\varnothing$

Juniven
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7

The boundary of any open (or closed) set is a closed set with empty interior. Any closed set with empty interior can be obtained in this way: if $F$ is closed and has empty interior, then $F$ is its own boundary, and also (of course) the boundary of its complement.

Martin Sleziak
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bof
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6

Here's one: $\{0\} \subset \mathbb{R}$

user3668561
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The fat Cantor set has not only empty interior but positive measure! This shows that having no interior is actually a very weak condition.

user21820
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