a contradiction in my proof to show that T must be the discrete topology if $f:(X,T)rightarrow (Bbb R,T_st)$, I dont understand .Show that d generates the discrete topologyContinuous functions in the product topology on $BbbR^BbbN$Show that $tau$ is not a topology on $BbbR$(Proof-verification) Constant maps are the only maps that are always continuousA Theorem Regarding Maps, the Trivial Topology and the Discrete TopologyClosure in the Discrete TopologyIs either $sigma=tau$ or $sigma$ is the is the discrete topology ? (True/false)A finite topological space is T1 if and only the topology is discreteHow can the preimage of a closed set for an open map be open?f nonsurjective map between two topological spaces; continuity
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a contradiction in my proof to show that T must be the discrete topology if $f:(X,T)rightarrow (Bbb R,T_st)$, I dont understand .
Show that d generates the discrete topologyContinuous functions in the product topology on $BbbR^BbbN$Show that $tau$ is not a topology on $BbbR$(Proof-verification) Constant maps are the only maps that are always continuousA Theorem Regarding Maps, the Trivial Topology and the Discrete TopologyClosure in the Discrete TopologyIs either $sigma=tau$ or $sigma$ is the is the discrete topology ? (True/false)A finite topological space is T1 if and only the topology is discreteHow can the preimage of a closed set for an open map be open?f nonsurjective map between two topological spaces; continuity
$begingroup$
I want to prove that T is the discrete topology on X if and only if every function $f:(X,T)rightarrow (Bbb R, T_st)$ is continuous.
Here's my attempt:
$(Rightarrow)$Assume That T is the discrete topology
using the theorem that a map is continuous if and only if the inverse of the function is open in the domain for every open set it maps in the codomain , I said that as all sets are open the discrete topology that obviously $f^-1(Bbb R)$ is open for all open sets in $(Bbb R, T_st)$
$(Leftarrow)$Now suppose f is continuous $Rightarrow$ for all $Usubset Bbb R$ the preimage $f^-1(U)$ is open in (X,T)
So we know that at least for open sets we'll map to open sets, we want to show that the same is true for closed sets too though
But heres where I get stuck, we assumed that f is continuous which means that $f^-1(Bbb R)$ is closed for every closed set in $(Bbb R, T_st)$ , but closed sets exist in $(Bbb R, T_st)$ which seems to imply that closed sets exist in $(X,T)$ which would mean that T couldn't be the discrete topology?
What am I missing here ?
general-topology proof-verification proof-writing
$endgroup$
add a comment |
$begingroup$
I want to prove that T is the discrete topology on X if and only if every function $f:(X,T)rightarrow (Bbb R, T_st)$ is continuous.
Here's my attempt:
$(Rightarrow)$Assume That T is the discrete topology
using the theorem that a map is continuous if and only if the inverse of the function is open in the domain for every open set it maps in the codomain , I said that as all sets are open the discrete topology that obviously $f^-1(Bbb R)$ is open for all open sets in $(Bbb R, T_st)$
$(Leftarrow)$Now suppose f is continuous $Rightarrow$ for all $Usubset Bbb R$ the preimage $f^-1(U)$ is open in (X,T)
So we know that at least for open sets we'll map to open sets, we want to show that the same is true for closed sets too though
But heres where I get stuck, we assumed that f is continuous which means that $f^-1(Bbb R)$ is closed for every closed set in $(Bbb R, T_st)$ , but closed sets exist in $(Bbb R, T_st)$ which seems to imply that closed sets exist in $(X,T)$ which would mean that T couldn't be the discrete topology?
What am I missing here ?
general-topology proof-verification proof-writing
$endgroup$
1
$begingroup$
Why can't closed sets exist in the discrete topology?
$endgroup$
– bitesizebo
yesterday
$begingroup$
@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
$endgroup$
– Dog_69
yesterday
$begingroup$
@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
$endgroup$
– Dog_69
yesterday
add a comment |
$begingroup$
I want to prove that T is the discrete topology on X if and only if every function $f:(X,T)rightarrow (Bbb R, T_st)$ is continuous.
Here's my attempt:
$(Rightarrow)$Assume That T is the discrete topology
using the theorem that a map is continuous if and only if the inverse of the function is open in the domain for every open set it maps in the codomain , I said that as all sets are open the discrete topology that obviously $f^-1(Bbb R)$ is open for all open sets in $(Bbb R, T_st)$
$(Leftarrow)$Now suppose f is continuous $Rightarrow$ for all $Usubset Bbb R$ the preimage $f^-1(U)$ is open in (X,T)
So we know that at least for open sets we'll map to open sets, we want to show that the same is true for closed sets too though
But heres where I get stuck, we assumed that f is continuous which means that $f^-1(Bbb R)$ is closed for every closed set in $(Bbb R, T_st)$ , but closed sets exist in $(Bbb R, T_st)$ which seems to imply that closed sets exist in $(X,T)$ which would mean that T couldn't be the discrete topology?
What am I missing here ?
general-topology proof-verification proof-writing
$endgroup$
I want to prove that T is the discrete topology on X if and only if every function $f:(X,T)rightarrow (Bbb R, T_st)$ is continuous.
Here's my attempt:
$(Rightarrow)$Assume That T is the discrete topology
using the theorem that a map is continuous if and only if the inverse of the function is open in the domain for every open set it maps in the codomain , I said that as all sets are open the discrete topology that obviously $f^-1(Bbb R)$ is open for all open sets in $(Bbb R, T_st)$
$(Leftarrow)$Now suppose f is continuous $Rightarrow$ for all $Usubset Bbb R$ the preimage $f^-1(U)$ is open in (X,T)
So we know that at least for open sets we'll map to open sets, we want to show that the same is true for closed sets too though
But heres where I get stuck, we assumed that f is continuous which means that $f^-1(Bbb R)$ is closed for every closed set in $(Bbb R, T_st)$ , but closed sets exist in $(Bbb R, T_st)$ which seems to imply that closed sets exist in $(X,T)$ which would mean that T couldn't be the discrete topology?
What am I missing here ?
general-topology proof-verification proof-writing
general-topology proof-verification proof-writing
edited yesterday
Robert Shore
2,234117
2,234117
asked yesterday
can'tcauchycan'tcauchy
1,009417
1,009417
1
$begingroup$
Why can't closed sets exist in the discrete topology?
$endgroup$
– bitesizebo
yesterday
$begingroup$
@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
$endgroup$
– Dog_69
yesterday
$begingroup$
@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
$endgroup$
– Dog_69
yesterday
add a comment |
1
$begingroup$
Why can't closed sets exist in the discrete topology?
$endgroup$
– bitesizebo
yesterday
$begingroup$
@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
$endgroup$
– Dog_69
yesterday
$begingroup$
@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
$endgroup$
– Dog_69
yesterday
1
1
$begingroup$
Why can't closed sets exist in the discrete topology?
$endgroup$
– bitesizebo
yesterday
$begingroup$
Why can't closed sets exist in the discrete topology?
$endgroup$
– bitesizebo
yesterday
$begingroup$
@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
$endgroup$
– Dog_69
yesterday
$begingroup$
@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
$endgroup$
– Dog_69
yesterday
$begingroup$
@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
$endgroup$
– can'tcauchy
yesterday
$begingroup$
@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
$endgroup$
– Dog_69
yesterday
$begingroup$
@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
$endgroup$
– Dog_69
yesterday
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The forward direction is incorrect as you state it: you need to note that if $X$ is discrete and $f: X to mathbbR$ is any function, then $f^-1[U]$ is open in $X$ for all open $U$ (not just $mathbbR$) in $mathbbR$. So $f$ is continuous.
If all real-valued functions on $X$ are continuous, let $A subseteq X$. The function $chi_A: X to mathbbR$ defined by $$chi_A(x)=begincases 1 & text if x in A\
0 & text if x notin A endcases$$
is continuous by assumption. This implies that $chi_A^-1[1] = A$ is closed as $1$ is closed in $mathbbR$. As $A$ is arbitary, all subsets of $X$ are closed, which implies that $X$ has the discrete topology.
$endgroup$
add a comment |
$begingroup$
Implication $implies$ is trivial, and you're not using the hypothesis for the converse in full force. You already know that every function $Xto Bbb R$ is continuous, so you're allowed to choose a function. Given $xin X$, our goal is to show that $x$ is open. Define the Dirac delta $delta_x:Xto Bbb R$ by $delta_x(x)=1$ and $0$ otherwise. By assumption $delta_x$ is continuous. So $x =delta_x^-1((1/2,+infty))$ is open.
$endgroup$
$begingroup$
I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
$endgroup$
– can'tcauchy
yesterday
$begingroup$
It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
$endgroup$
– Ivo Terek
yesterday
add a comment |
$begingroup$
For ($Leftarrow$) you could also use contraposition. Assume T is not the discrete topology then there exists a subset $Usubset X$, $X-Unotin T$ (U is not closed). Find a function that sends U to a closed subset of $mathbbR,$ for example:
$$f(x)=begincases
1& if& xin X-U \
0 &if& xin U
endcases
$$
Then since $f(U)=0$ is closed in $(mathbbR,T_st)$ and $f^-1(0)=U$ is not closed in $(X,T)$ it follows that $f$ is not continious.
New contributor
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3 Answers
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3 Answers
3
active
oldest
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active
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votes
$begingroup$
The forward direction is incorrect as you state it: you need to note that if $X$ is discrete and $f: X to mathbbR$ is any function, then $f^-1[U]$ is open in $X$ for all open $U$ (not just $mathbbR$) in $mathbbR$. So $f$ is continuous.
If all real-valued functions on $X$ are continuous, let $A subseteq X$. The function $chi_A: X to mathbbR$ defined by $$chi_A(x)=begincases 1 & text if x in A\
0 & text if x notin A endcases$$
is continuous by assumption. This implies that $chi_A^-1[1] = A$ is closed as $1$ is closed in $mathbbR$. As $A$ is arbitary, all subsets of $X$ are closed, which implies that $X$ has the discrete topology.
$endgroup$
add a comment |
$begingroup$
The forward direction is incorrect as you state it: you need to note that if $X$ is discrete and $f: X to mathbbR$ is any function, then $f^-1[U]$ is open in $X$ for all open $U$ (not just $mathbbR$) in $mathbbR$. So $f$ is continuous.
If all real-valued functions on $X$ are continuous, let $A subseteq X$. The function $chi_A: X to mathbbR$ defined by $$chi_A(x)=begincases 1 & text if x in A\
0 & text if x notin A endcases$$
is continuous by assumption. This implies that $chi_A^-1[1] = A$ is closed as $1$ is closed in $mathbbR$. As $A$ is arbitary, all subsets of $X$ are closed, which implies that $X$ has the discrete topology.
$endgroup$
add a comment |
$begingroup$
The forward direction is incorrect as you state it: you need to note that if $X$ is discrete and $f: X to mathbbR$ is any function, then $f^-1[U]$ is open in $X$ for all open $U$ (not just $mathbbR$) in $mathbbR$. So $f$ is continuous.
If all real-valued functions on $X$ are continuous, let $A subseteq X$. The function $chi_A: X to mathbbR$ defined by $$chi_A(x)=begincases 1 & text if x in A\
0 & text if x notin A endcases$$
is continuous by assumption. This implies that $chi_A^-1[1] = A$ is closed as $1$ is closed in $mathbbR$. As $A$ is arbitary, all subsets of $X$ are closed, which implies that $X$ has the discrete topology.
$endgroup$
The forward direction is incorrect as you state it: you need to note that if $X$ is discrete and $f: X to mathbbR$ is any function, then $f^-1[U]$ is open in $X$ for all open $U$ (not just $mathbbR$) in $mathbbR$. So $f$ is continuous.
If all real-valued functions on $X$ are continuous, let $A subseteq X$. The function $chi_A: X to mathbbR$ defined by $$chi_A(x)=begincases 1 & text if x in A\
0 & text if x notin A endcases$$
is continuous by assumption. This implies that $chi_A^-1[1] = A$ is closed as $1$ is closed in $mathbbR$. As $A$ is arbitary, all subsets of $X$ are closed, which implies that $X$ has the discrete topology.
answered yesterday
Henno BrandsmaHenno Brandsma
112k348120
112k348120
add a comment |
add a comment |
$begingroup$
Implication $implies$ is trivial, and you're not using the hypothesis for the converse in full force. You already know that every function $Xto Bbb R$ is continuous, so you're allowed to choose a function. Given $xin X$, our goal is to show that $x$ is open. Define the Dirac delta $delta_x:Xto Bbb R$ by $delta_x(x)=1$ and $0$ otherwise. By assumption $delta_x$ is continuous. So $x =delta_x^-1((1/2,+infty))$ is open.
$endgroup$
$begingroup$
I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
$endgroup$
– can'tcauchy
yesterday
$begingroup$
It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
$endgroup$
– Ivo Terek
yesterday
add a comment |
$begingroup$
Implication $implies$ is trivial, and you're not using the hypothesis for the converse in full force. You already know that every function $Xto Bbb R$ is continuous, so you're allowed to choose a function. Given $xin X$, our goal is to show that $x$ is open. Define the Dirac delta $delta_x:Xto Bbb R$ by $delta_x(x)=1$ and $0$ otherwise. By assumption $delta_x$ is continuous. So $x =delta_x^-1((1/2,+infty))$ is open.
$endgroup$
$begingroup$
I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
$endgroup$
– can'tcauchy
yesterday
$begingroup$
It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
$endgroup$
– Ivo Terek
yesterday
add a comment |
$begingroup$
Implication $implies$ is trivial, and you're not using the hypothesis for the converse in full force. You already know that every function $Xto Bbb R$ is continuous, so you're allowed to choose a function. Given $xin X$, our goal is to show that $x$ is open. Define the Dirac delta $delta_x:Xto Bbb R$ by $delta_x(x)=1$ and $0$ otherwise. By assumption $delta_x$ is continuous. So $x =delta_x^-1((1/2,+infty))$ is open.
$endgroup$
Implication $implies$ is trivial, and you're not using the hypothesis for the converse in full force. You already know that every function $Xto Bbb R$ is continuous, so you're allowed to choose a function. Given $xin X$, our goal is to show that $x$ is open. Define the Dirac delta $delta_x:Xto Bbb R$ by $delta_x(x)=1$ and $0$ otherwise. By assumption $delta_x$ is continuous. So $x =delta_x^-1((1/2,+infty))$ is open.
answered yesterday
Ivo TerekIvo Terek
46.4k954142
46.4k954142
$begingroup$
I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
$endgroup$
– can'tcauchy
yesterday
$begingroup$
It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
$endgroup$
– Ivo Terek
yesterday
add a comment |
$begingroup$
I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
$endgroup$
– can'tcauchy
yesterday
$begingroup$
It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
$endgroup$
– Ivo Terek
yesterday
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I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
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– can'tcauchy
yesterday
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I have a few questions that will probably sound silly to you but I want to better understand what you're saying , 1) if $delta_x$ maps any point in x to zero or one then the codomain will be 1,0, but then why can we choose elements outside of this set for our inverse map ? . 2) so we can choose any arbitrary function because f was arbitrary , sure. but how is it that by selecting one function we've shown it must be true for all functions ? 3) why was it I seemed to be getting a contradiction ?
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– can'tcauchy
yesterday
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It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
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– Ivo Terek
yesterday
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It does not map any point in $X$ to zero. I fixed a point $x in X$ to begin with, the goal is to prove that this particular singleton is open, and so we choose a suitable continuous function for that. About 2 and 3, it looks to me that your problem here is not topology, but logic. When you want to prove an "if and only if" statement, you have two separate implications to prove. If you assume that the topology on $X$ is the discrete and prove that every $Xto Bbb R$ is continuous, great. You're done with this implication. Now you have to prove something else, with new hypotheses.
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– Ivo Terek
yesterday
add a comment |
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For ($Leftarrow$) you could also use contraposition. Assume T is not the discrete topology then there exists a subset $Usubset X$, $X-Unotin T$ (U is not closed). Find a function that sends U to a closed subset of $mathbbR,$ for example:
$$f(x)=begincases
1& if& xin X-U \
0 &if& xin U
endcases
$$
Then since $f(U)=0$ is closed in $(mathbbR,T_st)$ and $f^-1(0)=U$ is not closed in $(X,T)$ it follows that $f$ is not continious.
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add a comment |
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For ($Leftarrow$) you could also use contraposition. Assume T is not the discrete topology then there exists a subset $Usubset X$, $X-Unotin T$ (U is not closed). Find a function that sends U to a closed subset of $mathbbR,$ for example:
$$f(x)=begincases
1& if& xin X-U \
0 &if& xin U
endcases
$$
Then since $f(U)=0$ is closed in $(mathbbR,T_st)$ and $f^-1(0)=U$ is not closed in $(X,T)$ it follows that $f$ is not continious.
New contributor
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add a comment |
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For ($Leftarrow$) you could also use contraposition. Assume T is not the discrete topology then there exists a subset $Usubset X$, $X-Unotin T$ (U is not closed). Find a function that sends U to a closed subset of $mathbbR,$ for example:
$$f(x)=begincases
1& if& xin X-U \
0 &if& xin U
endcases
$$
Then since $f(U)=0$ is closed in $(mathbbR,T_st)$ and $f^-1(0)=U$ is not closed in $(X,T)$ it follows that $f$ is not continious.
New contributor
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For ($Leftarrow$) you could also use contraposition. Assume T is not the discrete topology then there exists a subset $Usubset X$, $X-Unotin T$ (U is not closed). Find a function that sends U to a closed subset of $mathbbR,$ for example:
$$f(x)=begincases
1& if& xin X-U \
0 &if& xin U
endcases
$$
Then since $f(U)=0$ is closed in $(mathbbR,T_st)$ and $f^-1(0)=U$ is not closed in $(X,T)$ it follows that $f$ is not continious.
New contributor
New contributor
answered yesterday
SeriousSySeriousSy
1
1
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New contributor
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Why can't closed sets exist in the discrete topology?
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– bitesizebo
yesterday
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@bitesizebo from wolfram mathworld : A topology is given by a collection of subsets of a topological space . The smallest topology has two open sets, the empty set and the set itself. The largest topology contains all subsets as open sets, and is called the discrete topology. In particular, every point in X is an open set in the discrete topology.
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– can'tcauchy
yesterday
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@can'tcauchy Yes. But if every set is open, then, every set is closed. Just take complements.
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– Dog_69
yesterday
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@Dog_69 okay I understand what you mean , so then could I say as T is discrete all its sets are both open and closed which then means that all inverse images of f will map closed sets to closed sets and open to open because all sets in T are clopen ? I'm still a bit confused as how to show the converse though
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– can'tcauchy
yesterday
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@can'tcauchy Yes, exactly. This why functions from discrete spaces are open always.
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– Dog_69
yesterday