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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?

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    $\begingroup$ What are the open sets of a subspace topology? You have to show that every subset of the subspace is open. $\endgroup$ – BrianO Dec 7 '15 at 0:12
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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.

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  • $\begingroup$ 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 Dec 7 '15 at 0:31
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    $\begingroup$ @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 Dec 7 '15 at 0:47

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