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Consider a space $Y = (\mathbb R,\mathcal F)$, where $U$ is element of $\mathcal F$ iff $U$ is empty or $U$ contains the number $1$. (It can be proven that $Y$ is topological space).

Let $y_1,y_2$ be element of $Y$. Why is this function continuous?

$$ \begin{aligned} f \colon [0,1] &\to Y \\ t &\mapsto \left\{ \begin{aligned} y_1 \quad &\text{if}~ t = 0 \\ 1 \quad &\text{if}~ 0 < t < 1 \\ y_2 \quad &\text{if}~ t = 1 \\ \end{aligned} \right. \end{aligned}$$

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where do you think you have to seriously check for continuity? –  Praphulla Koushik Dec 7 '13 at 7:36
    
I dont understand how I can take a pull back (PreImage) and show that they are still open. –  da_elysian_fields Dec 7 '13 at 7:42
    
look for the answer below... it is more informative.. –  Praphulla Koushik Dec 7 '13 at 7:44
    
Note that it is continuous for the usual topology on $[0,1]$, not for the subspace topology of $[0,1] \subset Y$. –  Pece Dec 7 '13 at 7:44
    
Well that clears the question better but what I dont understand how do we take preimage of open set? I would be thankful if you could elaborate with an example. –  da_elysian_fields Dec 7 '13 at 8:11
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2 Answers 2

up vote 2 down vote accepted

the pre-image of any nonempty open set in $Y$ must be one of $[0,1), (0,1)$ or $(0,1]$, since all of these are open in the usual topology for the closed unit interval $I$, the pre-image of any open set in $Y$ is open in $I$. but this is the condition for continuity.

(a few remarks added in response to comments and OP's request for clarification)

in reading math we frequently see confessions like: "this abuse of notation should cause no difficulties..." hmmm.

one of the most frequent abuses of notation is the use, for a mapping $f$ of the symbol $f^{-1}$. this has a clear meaning only when $f$ is both surjective and injective. otherwise the inverse function exponent $^{-1}$ conveniently avoids mentioning the fact, which might easily bewilder those new to the subject, that we are in fact dealing with a contravariant functor from the category of sets to its subcategory of powersets.

with the usual notation $\mathfrak{P}(A)$ for the operation of forming the powerset, then if we have a function $f:A \rightarrow B$ then under the action of the (more familiar) covariant powerset functor we have:

$$ \mathfrak{P}(f): \mathfrak{P}(A) \rightarrow \mathfrak{P}(B) \\ $$ defined by $$ \forall \mathbb{a} \in \mathfrak{P}(\mathbb{a}),\; \mathfrak{P}(f)(\mathbb{a}) = \bigcup_{a \in \mathbb{a}} \{f(a)\} $$ this is very natural for our minds to think about, so the categoric underpinning is left aside as excess baggage. this shows, for example, in refusing to use a notation which clearly distinguishes $f$ and $\mathfrak{P}(f)$ or in the way the set-forming operator $\{\dots\}$ would usually be omitted from the union on the right. the identification with an element $x$ of $A$ with the corresponding singleton set $\{a\}$ in $\mathfrak{P}(A)$ is easy and does not cause much trouble at an elementary level.

the contravariant powerset functor, which we may denote here by $\mathfrak{P}^{-1}$ for emphasis, takes each set to its powerset as before, but its effect on functions is different. in fact we have: $$\mathfrak{P}^{-1}(f):\mathfrak{P}(B) \rightarrow \mathfrak{P}(A)$$ a reversal of direction flagged by the prefix contra-. its action is defined by: $$ \forall \mathbb{b} \in \mathfrak{P}(B),\; \mathfrak{P}^{-1}(f)(\mathbb{b}) = \bigcup_{b \in \mathbb{b}} \{a \in A \mid f(a)=b\} $$ so typically we use the notation $f^{-1}$ as a convenient abbreviation for $\mathfrak{P}^{-1}(f)$.

as an example of the utility of this more precise way of looking at the matter, note that we may say $f:A \rightarrow B $ is surjective iff $$ \forall \mathbb{b} \in \mathfrak{P}(B), \;f^{-1}(\mathbb{b}) = \emptyset \rightarrow \mathbb{b} = \emptyset $$

also note that a topology on a set $A$ is actually just a single element of the set $\mathfrak{P}^2(A)$, as should be clear from its description as a family of subsets of $A$.

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Thanks david for your response but could you tell me with an example or two why open set in Y must be one of [0,1),(0,1) or (0,1]? –  da_elysian_fields Dec 7 '13 at 8:15
    
I meant why " pre-image of any nonempty open set in Y must be one of [0,1),(0,1) or (0,1]"? –  da_elysian_fields Dec 7 '13 at 8:22
    
@da_elysian_fields: If $V \subseteq Y$ is a nonempty open set, there are four further cases: (1) $y_1,y_2 \in V$; (2) $y_1 \in V$, $y_2 \notin V$; (3) $y_1 \notin V$, $y_2 \in V$; and (4) $y_1,y_2 \notin V$. Check these cases one-by-one, and remember that $1 \in V$. –  Arthur Fischer Dec 7 '13 at 8:27
    
You are missing one case for $f^{-1} [ V ]$ for nonempty open $V$ in your answer. –  Arthur Fischer Dec 7 '13 at 8:28
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There is an equivalent definition of continuity: a function $f: X\to Y$ between topological spaces is continuous at $x$ if there for any neighbourhood $V$ of $f(x)$ exists a neighbourhood $U$ of $x$ such that $f(U) \subseteq V$. The function is continuous if it is continuous at every $x\in X$.

In your case:

$t\neq 0,1:$ then $f(t) = 1$. Around any such $t$ we choose $U=(0,1)$ and $f(U)=1$.

$t=0:$ choose any open set $V$ around $y_1$. Then $f([0,1))\subseteq V$ since every open set contains $1$.

$t=1:$ basically the same as for $t=0$.

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