Is 1 in the Closure of (2,3] in the Standard Topology on the Real Numbers?

[SYoungblood]

Homework Statement

Hello All, I am experiencing Adventures in Topology. So far, so good, but I have an issue here.

In the topological space (Real #s, U), show that 1 is not an element of Cl((2,3]).

Homework Equations

The closed subsets of our topological space are the converses of the given set. Over the set of real numbers, in an open topology, we have a converse of (-inf, 2] u (3,inf)

The Attempt at a Solution

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By that definition, 1 is most certainly an element of Cl((2,3]), and I am simply not seeing otherwise. Any thoughts?

Thank you,

SY

 

[fresh_42]

What do you mean by #s and U? What exactly are your open sets, or likewise closed sets? Does Cl mean closure? In addition I assume by converse you mean the complement, but I'm not sure.

How is the closure of a set defined?

 

[SYoungblood]

What do you mean by #s and U? What exactly are your open sets, or likewise closed sets? Does Cl mean closure? In addition I assume by converse you mean the complement, but I'm not sure.

How is the closure of a set defined?

Real #s is the best notation I could think of for the set of Real Numbers. The set (2,3] is neither open nor closed under open topology, which I noted as U, in the fashion of my test. Cl is a closed space. As an example, in the space (Real #s, U), Cl((0,1)) = [0,1], the compliment of the interval (0,1) over the set of real numbers.

With that in mind, again, I am still not seeing how to exclude 1 from the interval of Cl ((2,3]).

 

[fresh_42]

To get the wording straight. You consider the topological space $\mathbb{R}$ with the standard topology, which is induced by the Euclidean metric, which measures distances as $d(x,y) = |x-y|$. Now $1 \notin (2,3]$ and you want to show, whether $1$ can be in the closure of this interval or not. Here is the essential part of your conclusion, because it requires a definition of the closure of a set. This can be done in several ways, purely topological or by means of the metric. But you have said, that $Cl((2,3])=[2,3]$ according to your example. So where is the problem?

Otherwise, you have to say, which definition of a closure of a set you want to use, before we can achieve a formal proof. What does your book say?