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I want to know why a discrete topological space is discrete"? I looked up some references books, but found nothing about this question. In my opinion,discrete" is nearly related to ``discontinuous". In order to answer this question, I figure out a corresponding assertion:

Let $f$ be a non-constant map from topological space $ X$ to discrete space $Y.$ Then $f$ is continuous, if and only if $X$ is a discrete space.

The sufficiency part of the foregoing assertion is trivial. But I do not know if its necessity part is valid. Can anyone help me?

azc
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  • A better version would read: Any function $f:X\to Y$ to any topological space $Y$ is continuous, iff $X$ is discrete. In general Discrete Math is in some vague sense the opposite of Continuous Math (analysis and such). Not unexpectedly that is one sweeping generalization. Methods from one side of the aisle are applied to questions from the other at leisure. – Jyrki Lahtonen Apr 13 '15 at 09:35
  • @JyrkiLahtonen. I do not know why you modified my assertion like that. I want to give some conditions which yields that, even ordinary continuous function, like $f(x)=\sin(x) $ from $\mathbb{R} $ to $\mathbb{R}$ will be disretizied to be discontinuous, if we modified the topology of the codomain. – azc Apr 14 '15 at 04:47
  • I was looking for a modification that gives a true statement :-) – Jyrki Lahtonen Apr 14 '15 at 06:06
  • Your latest addition (where the assumptions only mention a GIVEN function $f$) only means that $X$ cannot be connected. You need to quantify over the set of ALL functions to force $X$ to be discrete. – Jyrki Lahtonen Apr 14 '15 at 06:08
  • For example, let $X=[0,1]\cup [2,3]$ with the usual topology, and $Y={0,1}$ with discrete topology. The non-constant function $f$ that maps all of $[0,1]$ to $0$ and all of $[2,3]$ to $1$ is continuous. Yet $X$ is not discrete. – Jyrki Lahtonen Apr 14 '15 at 06:12

2 Answers2

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The statement you wrote is false, since a constant $f$ is continuous, no matter what topology you apply to $Y$.

5xum
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Maybe because such a space (X,T) is metrizable by the discrete metric and then the basis can be taken as the collection of all open balls B(x,1/2) ={x} for all x in X.

Adelafif
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