Topology: Continuity - discrete and indiscrete spaces

[BadBe...@googlemail.com]

Hello,
I just started my jourey into Topology and stumpled on the following:
(i) Every function from a discrete space into any top. space is
continuous.
(ii) Every function from any top space into an indiscrete space is
continuous.
Now, unfortunately I am not quite sure why that is. I hope some
experienced traveller can help me out with an explanation!
Bst Regards BBH

[Tonico]

On Jul 4, 2:37 pm, "BadBedH...@googlemail.com"

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Let X be a discrete space, i.e.: every subset of X is open, and let
now f: X --> H be any function from X to any top. space H.
Let U be open in H: is f^(-1)(U) open in X? Of course it is...can you
see why?

If now Y is an indiscrete top. space, then its only open subsets are
the empty set and Y itself. Any function f: Z --> Y from any top.
space Z will have only two open sets in Y to check on: so, what are
the inverse images of the empty set and of the whole space Y?

Regards
Tonio

[William Elliot]

On Fri, 4 Jul 2008, BadBe...@googlemail.com wrote:

> I just started my journey into Topology and stumped on the following:

> (i) Every function from a discrete space into any top. space is
> continuous.
> (ii) Every function from any top space into an indiscrete space is
> continuous.

> Now, unfortunately I am not quite sure why that is.

Let us first look at the topological definition of continuous.

f:X -> Y is continuous when
for all open U subset Y, f^-1(U) open subset X.

This is the definition that requires study and to assure yourself
that in encompasses the definition of continuous of metric spaces
and the usual definition of continuous in analysis.

Another definition with the same need for study.
f:X -> Y is continuous at a when for all open V nhood f(a),
some open U nhood x with f(U) subset V.

Theorem. f:X -> Y is continuous iff for all a in X,
f is continuous at a.

Anyway, let X be a discrete space.
If U open subset Y, then of course f^-1(U) is open.
Thus immediately f is continuous.

If Y is the indiscrete space, then the open sets of Y are
nulset and Y only. Since f^-1(nulset) = nulset and
f^-1(Y) = X are both open, f is continuous.

[BadBe...@googlemail.com]

Thanks for your answers!