Closure and completeness

[davidebianco1]

Hello Everyone,

first of all my apologies, may be my question is too stupid for a forum on Topology and Geometry, but it's something I was thinking about for a while without getting an answer : What's the actual difference between a Close set and a Complete set? I mean : from an algebraic point of view, we say that a Close set contains the limits of its successions, whereas a complete set contains the limits of its Cauchy successions. But we can show that a succession is convergent if and only if it is a Cauchy succession, so the two concepts should be equivalent, and I don't see the difference between closure and completeness. From a topological point of view, we define a close set as the complement of an open one, but at this point I don't see the connection with the completeness property. I started to taught the actual difference could be in passing from infinite dimension to finite dimension, but I am not able to see clearly why. Can anyone help me?
Sorry again if the question is not so relevant. Cheers,


Davide

 

[yyat]

Hi Davide,

Closedness is a property a subset of a any topological space can have. It is "extrinsic" because it depends on which space you are considering it to be a subset of, for example the interval (0,1] is closed in (0,2] but not in [0,2] (with the usual topology).

Completeness, on the other hand, makes sense only for metric spaces (or so-called uniform spaces). It is an intrinsic property.

Sometimes however, the two notions coincide, for example for subvectorspaces of a Banach-space.

 

[Edwin]

Yyat,

You had mentioned that (0, 1] is an open set in the topological space (0, 2] with the topology on (0, 2] inherited from the topology on R.

Would one way to prove that (0 , 1] is open in (0, 2] be to find a homeomorphism from
(0, 2] onto (0, 1], then conclude that (0, 1] is open in (0, 2] since (0, 2] is open by definition of a topological space, and homeomorphisms from topological spaces to topological spaces maps open sets to open sets?

For this particular case, I am thinking the proof could be done by making use of the homeomorphism of scalar multiplication:

x->bx, with b non-zero and real, and then noting that (0, 1] = 1/2*(0, 2].

What are your thoughts?

 

[davidebianco1]

Hello again guys,

so, If I have understood well, the question is that, in principle, closure and completeness are concepts different in nature, and it can just happen they do coincide somewhere? (Actually, I didn,t get the example about the interval on R). Many thanks for replying and Cheers


Davide

 

[yyat]

Yyat,

You had mentioned that (0, 1] is an open set in the topological space (0, 2] with the topology on (0, 2] inherited from the topology on R.

(0,1] is closed in (0,2]!

Would one way to prove that (0 , 1] is open in (0, 2] be to find a homeomorphism from
(0, 2] onto (0, 1], then conclude that (0, 1] is open in (0, 2] since (0, 2] is open by definition of a topological space, and homeomorphisms from topological spaces to topological spaces maps open sets to open sets?

This only proves that (0,1] is open in (0,1] itself.

For this particular case, I am thinking the proof could be done by making use of the homeomorphism of scalar multiplication:

x->bx, with b non-zero and real, and then noting that (0, 1] = 1/2*(0, 2].

What are your thoughts?

I would prove that (0,1] is closed in (0,2] as follows:
(1,3) is open in R.
By definition of the induced topology (1,2]=(1,3)$$\cap$$(0,2] is open in (0,2]. (If B is a subset of A, the induced topology on B is given by sets of the form B$$\cap$$U, where U is any open set in A).
Therefore, (0,1] is the complement of an open set (in (0,2]), hence closed by definition.

 

[Edwin]

Thank you! Sorry about the inaccurate statements: when I went to post, I got the ideas mixed up in my mind.

The induced topology looks like a real convenient way to prove openness: if you have a topological space X, and C is a subset of B is a subset of X, all you have to do to prove that the subset set C of B is open in B is to show that C can be written as the intersection of B with some open set V in X.

A homeomorphism from a topological space X to a topological space Y maps subsets of X that are open in X to subsets of Y that are open in Y.

So, if B is a subset of A, and A and B are homeomorphic, is it still possible that an open set in B is not open in A?

 

[yyat]

So, if B is a subset of A, and A and B are homeomorphic, is it still possible that an open set in B is not open in A?

Yes. For instance take A=[0,2], B=[0,1], then B is open in B but not in A.

Maybe the following fact will clarify things: If A is an open subset of a topological space X and is given the induced topology, then any subset B of A that is open in A is also open in X. (Try showing this!)

 

[Edwin]

Yyat wrote: If A is an open subset of a topological space X and is given the induced topology, then any subset B of A that is open in A is also open in X. (Try showing this!)

I appreciate all of your help, and knowledge.

I'll give it a try.

Please let me know if I make a mistake in the proof.

Proof:

Let T be a topology on X. Then since A is open in X, then A is contained in T. Let K be the induced topology on A given by defining

$$K = \{A \cap V$$: $$V \in T \}$$.

If B is a subset of A, and B is open in A, then B is contained in K, and it follows by the definition of K that

$$B = A \cap V$$ for some set V in T.

Since this V is contained in T, and A is contained in T, then by the axioms of a topological space (which state that topologies are closed under finite intersection), we have

$$A \cap V$$ is contained in T.

That is, $$A \cap V$$ is an open set in X.

Since $$B = A \cap V$$,

then by the substitution principle, B is open in X, and the proof is complete.

 

[de_brook]

Your proof is correct. The induced topology K can be siad to be the relative toplolgy on A if A is considered as an open subspace of X.

 

[de_brook]

Is there a linear space in any union of linear subspaces is a subspace except the trivial subspaces? pls help

 

[de_brook]

Is there a linear space in which any union of linear subspaces is a subspace except the trivial subspaces? pls help

 

[Edwin]

I am not sure I am interpreting the question correctly. But as far as I know, given any linear space (vector space) W, the trivial linear subspaces of W are the origin {0} in W, and W itself.

Given any linear space X, it can be shown that the union of any of the following spaces

{{0}, X, Y}} where the trivial subspace {0} is the origin in X, and Y is any arbitrarily chosen subspace of X.

So, there exists no linear space in which any union of linear subspaces is a subspace except the trivial subspaces.

I have a quick: we know from set theory that the cardinality of the power set P(S) of any set S is strictly greater than the cardinality of S itself.

With that in mind, is it necessarily true that given any linear space (vector space) Z, the set Sub{Z} of all subspaces of Z is itself a subspace of Z?