2
$\begingroup$

This question actually has two parts:

a. Prove that every subspace of a topological space with the discrete topology has the discrete topology.

b. Prove that every subspace of a topological space with the trivial topology has the trivial topology.

If I can do part a, part b will be easy. What do I need to demonstrate, exactly, to show that a is true?

$\endgroup$
1
  • 2
    $\begingroup$ What are the open sets of a subspace topology? You have to show that every subset of the subspace is open. $\endgroup$
    – BrianO
    Commented Dec 7, 2015 at 0:12

1 Answer 1

5
$\begingroup$

Suppose we have a topological space $(X, \tau_X)$, where $\tau_X\subseteq 2^X$ is a topology on $X$. Then every subset $Y\subseteq X$ we can consider as a topological subspace of $(X, \tau_X)$ with induced topology $\tau_Y$. This topology constructs simply like this: $$ \tau_Y = \{U\cap Y:\ \ U\in\tau_X \} $$

So (a) if you have a topological space with the discrete topology $(X, 2^X)$ then for every $Y\subseteq X$ you'll have induced topology $\tau_Y = \{U\cap Y:\ \ U\in 2^X \}$ which is $2^Y$, just as was stated.

And (b) if you have a topological space with the antidiscrete (trivial) topology $(X, \{\emptyset, X\})$, then the induced topology on $Y$ will look like this $\tau_Y = \{U\cap Y:\ \ U\in \{\emptyset, X\} \}=\{\emptyset, Y\}$, just as expected.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer. I have a question - this is not the first time I've seen this notation $(X,2^X)$ on here, but I actually never saw this in my textbook. Maybe a stupid question, but what does this notation mean? $\endgroup$
    – Indigo
    Commented Dec 7, 2015 at 0:31
  • 3
    $\begingroup$ @Indigo, topological space is a pair of two things: a non-empty set (called carrier set, set of points) and a collection of subsets of the carrier set complied with topology axioms (called open sets). So topological space can be written as $(X, \tau)$, where $X$ is a carrier set and $\tau$ is a collection of subsets of the carrier set. $2^X$ denotes the set of all subsets of the set $X$. So, by definition, $\tau \subseteq 2^X$. For discrete topology by definition $\tau = 2^X$ $\endgroup$
    – Glinka
    Commented Dec 7, 2015 at 0:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .