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The Discrete Topology and the Discrete Metric Space ... Stromberg, Example 3.10 (b) ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Karl R. Stromberg's book: "An Introduction to Classical Real Analysis". ... ...

I am focused on Chapter 3: Limits and Continuity ... ...

I need help in order to fully understand Example 3.10 (b) on page 95 ... ...


Example 3.10 (b) reads as follows:




Stromberg - Example 3.10 (b) ...  .png




My question is as follows:


Stromberg says that if \(\displaystyle X\) is any set and \(\displaystyle \mathscr{T}\) is the family of all subsets of \(\displaystyle X\) ...

... then \(\displaystyle \mathscr{T}\) is nothing but the metric topology obtained from the discrete metric ...


Can someone demonstrate/explain exactly how/why this is true ...?




Help will be much appreciated ...

Peter




===================================================================================


Example 3.10 (b) above refers to Example 3.2 (a) ... ... so I am providing access to the same ... as follows:




Stromberg - Example 3.2 (a) ... Discrete Metric Space .png






It may help readers of the above post to have access to Stromberg's definition of a topological space ... so I am providing access to the same ... as follows:



Stromberg -  Defn 3.9  ... Defn of a Topological Space ... .png




Stromberg's definition of a topological space refers to Theorem 3.6 ... ... so I am providing access to the statement of the same ... as follows:



Stromberg - Statement of Theorem 3.6 ... .png




Stromberg's definition of a topological space also refers to Definition 3.3 ... ... so I am providing access to the same ... as follows:




Stromberg -  Defn 3.3  ... Defn of a Ball of Radius r with Center a  ... .png




Hope that helps ...

Peter
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
Stromberg says that if \(\displaystyle X\) is any set and \(\displaystyle \mathscr{T}\) is the family of all subsets of \(\displaystyle X\) ...

... then \(\displaystyle \mathscr{T}\) is nothing but the metric topology obtained from the discrete metric ...


Can someone demonstrate/explain exactly how/why this is true ...?
Let $\rho$ be the discrete metric on the set $X$, and let $x\in X$. Then $\rho(x,y) = 1$ whenever $y\ne x$. Therefore $B_1(x) = \{y\in X:\rho(x,y)<1\} = \{x\}$. In words, that says that $B_1(x)$ is an open set consisting of the single point $x$.

Now let $S$ be a subset of $X$. Then \(\displaystyle S = \bigcup_{x\in S}\{x\} = \bigcup_{x\in S}B_1(x)\). That is a union of open sets and is therefore open. Again putting it in words, this says that in the discrete metric every subset of $X$ is open.

So the topology obtained from the discrete metric (which by definition is the family of all open subsets of $X$) is the family of all subsets of $X$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Let $\rho$ be the discrete metric on the set $X$, and let $x\in X$. Then $\rho(x,y) = 1$ whenever $y\ne x$. Therefore $B_1(x) = \{y\in X:\rho(x,y)<1\} = \{x\}$. In words, that says that $B_1(x)$ is an open set consisting of the single point $x$.

Now let $S$ be a subset of $X$. Then \(\displaystyle S = \bigcup_{x\in S}\{x\} = \bigcup_{x\in S}B_1(x)\). That is a union of open sets and is therefore open. Again putting it in words, this says that in the discrete metric every subset of $X$ is open.

So the topology obtained from the discrete metric (which by definition is the family of all open subsets of $X$) is the family of all subsets of $X$.





Thanks for a most helpful post, Opalg ...

Just reflecting on what you have said ...

Thanks again ...

Peter